Use the quadratic equation to solve for the roots.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the values of the polynomial $\displaystyle y=ax^2+bx+c$ into the equation.

$\displaystyle y=2x^2-3x+20$

$\displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(20)}}{2(2)}$

Simplify the quadratic formula.

$\displaystyle x=\frac{+3\pm\sqrt{9-160}}{4} = \frac{3\pm\sqrt{-151}}{4}$

Since we have a negative discriminant, we will have complex roots even though there are no real roots.

The roots are:  $\displaystyle x=\frac{3\pm i\sqrt{151}}{4}$

One of the possible roots given is:  $\displaystyle \frac{3- i\sqrt{151}}{4}$

$\displaystyle 2x^2+3x-14$

There are multiple methods to solve quadratics. Use whichever is easiest for the problem.

Solve by factoring:

$\displaystyle (2x+7)(x-2)$

What numbers will multiply to get $\displaystyle 2x^2$ and what two numbers will multiply to get -14?

To solve for the roots, set the factors equal to 0 and solve:

$\displaystyle 2x+7=0\rightarrow \mathbf{x=-\frac{7}{2}}$

$\displaystyle (x-2)=0\rightarrow \boldsymbol{x=2}$

To find the roots, or solutions, of this quadratic equation, first factor it.

Recall that when a function is in the $\displaystyle ax^2+bx+c$ form, the factors of a and c when multiplied and added together must equal b.

First, identify factors of 6 that could equal -1. Y

ou have to think of your positive and negative signs here. Remember that $\displaystyle 3\cdot 2=6$ and you need to have one positive and one negative number to get -6.

Therefore, after factoring, you should get: 

$\displaystyle (x-3)(x+2)$.

Then, set each of those expressions equal to 0.

Therefore, your roots are:

$\displaystyle x=3,-2$.

Since this function is already factored and equal to 0, you can just set each expression equal to 0 to get your roots.

$\displaystyle 4x-2=0; x=\frac{1}{2}$

and

$\displaystyle x-3=0, x=3$.

The equation will need to be simplified to its standard form.

Simplify this equation by using the FOIL method to expand $\displaystyle (3-3x)^2$.

$\displaystyle (3-3x)(3-3x) = 3(3)+3(-3x)+(-3x)(3)+(-3x)(-3x)$

Simplify the terms.

$\displaystyle (3-3x)(3-3x) = 9-18x+9x^2$

The equation becomes:

$\displaystyle y=-2(9-18x+9x^2)-3x$

Distribute the negative two through the parentheses.

$\displaystyle y=-18+36x-18x^2-3x$

Combine like terms.

The equation in standard for becomes:  $\displaystyle y=-18x^2+33x-18$

The standard form for a polynomial is:  $\displaystyle y=ax^2+bx+c$

Write the quadratic equation.

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the coefficients corresponding to the equation in standard form.

$\displaystyle x=\frac{-33\pm\sqrt{33^2-4(-18)(-18)}}{2(-18)} =\frac{-33\pm\sqrt{-207}}{-36}$

Simplify the radical.  The roots will be imaginary.

$\displaystyle x=\frac{-33\pm\sqrt{-207}}{-36} = \frac{33}{36}\pm \frac{\sqrt{-1}\cdot \sqrt{9}\cdot \sqrt{23}}{36}$

Simplify the fractions and replace $\displaystyle \sqrt{-1}$ with $\displaystyle i$.

$\displaystyle x=\frac{11}{12}\pm \frac{3i\sqrt{23}}{36} = \frac{11}{12}\pm \frac{i\sqrt{23}}{12}$

The answer is:  $\displaystyle \frac{11}{12}- \frac{i\sqrt{23}}{12}$

You can solve this quadratic in many different ways, including by graphing or using the quadratic formula. This particular quadratic can be factored:

$\displaystyle 2x ^2 + 5 x - 12$ we're looking for 2 numbers that add to 5 and multiply to $\displaystyle -12 \cdot 2 = -24$

The numbers that work are 8 and -3:

$\displaystyle 2x^2 + 8x - 3x - 12$ continue factoring

$\displaystyle 2x(x+4) + -3 (x+4)$

The factors are $\displaystyle (x+4)$ and $\displaystyle (2x-3)$. Set each factor equal to zero:

$\displaystyle x+4 = 0$

$\displaystyle x=-4$

$\displaystyle 2x-3 = 0$

$\displaystyle 2x = 3$

$\displaystyle x=1.5$

You can solve this quadratic equation in many different ways, including by graphing or factoring. You can also use the quadratic formula:

$\displaystyle x = \frac{-17 \pm \sqrt{289 - 4(4)(-77)}}{2\cdot4} = \frac{-17 \pm \sqrt{1,521}}{8} = \frac{-17 \pm39}{8}$

This will give us two answers:

$\displaystyle \frac{-17 + 39 }{ 8 } = 2.75$

$\displaystyle \frac{-17-39}{8 } = -7$

This quadratic equation can be solved in several ways, including the quadratic equation or by graphing. It can also be solved by factoring:

For this quadratic, we are looking for 2 numbers that add to -19 and multiply to 6x10=60. The numbers satisfying these conditions are -4 and -15:

$\displaystyle 6x^2 -4x - 15x + 10$ continue factoring

$\displaystyle 2x(3x - 2 ) - 5 (3x - 2 )$

The factors are $\displaystyle (2x -5 )$ and $\displaystyle (3x - 2 )$. Set each equal to zero:

$\displaystyle 2x-5=0$

$\displaystyle 2x = 5$

$\displaystyle x=2.5$

$\displaystyle 3x-2 = 0$

$\displaystyle 3x = 2$

$\displaystyle x=\frac{2}{3}$

This quadratic can be solved in several different ways, including by graphing or factoring. You can also use the quadratic formula:

$\displaystyle x= \frac{ 5 \pm \sqrt{ 25 - 4(2)(-25)}}{2 \cdot 2 } = \frac{5 \pm \sqrt{225}}{4} = \frac{5 \pm 15}{4}$

This gives us two answers:

$\displaystyle \frac{5 + 15 } {4} = \frac{20}{4} = 5$

$\displaystyle \frac{5-15}{4} = \frac{-10}{4} = -2.5$

Use the quadratic equation to determine roots of a parabolic function.

The quadratic formula is:

$\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equation $\displaystyle y=4x^2-4x-3$ in the form of $\displaystyle y=ax^2+bx+c$.

Substitute the known coefficients.

$\displaystyle x= \frac{-(-4)\pm\sqrt{(-4)^2-4(4)(-3)}}{2(4)}$

Simplify the equation.

$\displaystyle x= \frac{4\pm\sqrt{16+48}}{8} = \frac{4\pm\sqrt{64}}{8} = \frac{4\pm8}{8}$

The fraction can be broken into two.

$\displaystyle \frac{4+8}{8} = \frac{12}{8}=\frac{3}{2}$

$\displaystyle \frac{4-8}{8} = \frac{-4}{8}= -\frac{1}{2}$

Either $\displaystyle -\frac{1}{2}$ or $\displaystyle \frac{3}{2}$ is a possible root.

The answer is: $\displaystyle -\frac{1}{2}$