First, subtract 4 from both sides:

\(\displaystyle \dpi{100} 3x^{2}+4-4=31-4\)

\(\displaystyle \dpi{100} 3x^{2}=27\)

Next, divide both sides by 3:

\(\displaystyle \dpi{100} \frac{3x^{2}}{3}=\frac{27}{3}\)

\(\displaystyle \dpi{100} x^{2}=9\)

Now take the square root of both sides:

\(\displaystyle \dpi{100} \sqrt{x^{2}}=\sqrt{9}\)

\(\displaystyle \dpi{100} x=\pm 3\)

\(\displaystyle \dpi{100} \frac{x}{3}+22 = 45\)

\(\displaystyle \dpi{100} \frac{x}{3}+22-22 = 45-22\)

\(\displaystyle \dpi{100} \frac{x}{3} = 23\)

\(\displaystyle \dpi{100} (3)\cdot \frac{x}{3} = 23\cdot (3)\)

\(\displaystyle \dpi{100} x=69\)

Rewrite this as a compound statement and solve each separately:

\(\displaystyle | 3x + 9| = 24\)

\(\displaystyle 3x + 9 = - 24 \textrm{ or }3x + 9 = 24\)

\(\displaystyle 3x + 9 = - 24\)

\(\displaystyle 3x + 9 -9= - 24-9\)

\(\displaystyle 3x = - 33\)

\(\displaystyle 3x\div 3 = - 33 \div 3\)

\(\displaystyle x = -11\)

\(\displaystyle 3x + 9 = 24\)

\(\displaystyle 3x + 9 -9= 24-9\)

\(\displaystyle 3x = 15\)

\(\displaystyle 3x\div 3 = 15 \div 3\)

\(\displaystyle x = 5\)

FOIL each of the two expressions, then solve:

\(\displaystyle \left (x -4 \right )\left( x +3 \right )= \left (x-2 \right) \left ( x-1\right )\)

\(\displaystyle x ^{2} + 3 \cdot x -4\cdot x -4 \cdot 3 = x^{2} -1 \cdot x - 2 \cdot x + 2\cdot1\)

\(\displaystyle x ^{2} + 3 x -4 x -12 = x^{2} -1 x - 2x + 2\)

\(\displaystyle x ^{2} - x -12 = x^{2} - 3x + 2\)

\(\displaystyle x ^{2} - x -12 - x^{2} = x^{2} - 3x + 2 - x^{2}\)

\(\displaystyle - x -12= - 3x + 2\)

Solve the resulting linear equation:

\(\displaystyle - x -12 +3x + 12 = - 3x + 2+3x+12\)

\(\displaystyle 2x =14\)

\(\displaystyle 2x\div 2 =14 \div 2\)

\(\displaystyle x = 7\)

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

\(\displaystyle x^{2} -10x = 24\)

\(\displaystyle x^{2} -10x -24= 24-24\)

\(\displaystyle x^{2} -10x -24= 0\)

Now factor the quadratic expression \(\displaystyle x^{2} -10x -24\) into two binomial factors \(\displaystyle (x + ?)(x + ?)\), replacing the question marks with two integers whose product is \(\displaystyle -24\) and whose sum is \(\displaystyle -10\). These numbers are \(\displaystyle -12,2\), so:

\(\displaystyle (x -12)(x + 2) = 0\)

\(\displaystyle x-12 = 0 \Rightarrow x = 12\)

or

\(\displaystyle x+2 = 0 \Rightarrow x = -2\)

The solution set is \(\displaystyle \left \{ -2,12\right \}\).

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

\(\displaystyle \left ( x-3\right )(2x-1) = 18\)

\(\displaystyle x\cdot 2x -x\cdot 1 -3 \cdot 2x +3\cdot 1= 18\)

\(\displaystyle 2x^{2} -x -6x +3= 18\)

\(\displaystyle 2x^{2} -7x +3= 18\)

\(\displaystyle 2x^{2} -7x +3-18= 18 -18\)

\(\displaystyle 2x^{2} -7x -15= 0\)

Use the \(\displaystyle ac\) method to factor the quadratic expression \(\displaystyle 2x^{2} -7x -15\); we are looking to split the linear term by finding two integers whose sum is \(\displaystyle -7\) and whose product is \(\displaystyle 2 (-15) = -30\). These integers are \(\displaystyle -10,3\), so:

\(\displaystyle (2x^{2} -10x)+(3x -15)= 0\)

\(\displaystyle 2x (x -5)+3(x -5)= 0\)

\(\displaystyle \left (2x +3 \right) \left ( x-5 \right )= 0\)

Set each expression equal to 0 and solve:

\(\displaystyle 2x +3= 0\)

\(\displaystyle 2x = -3\)

\(\displaystyle x = -\frac{3}{2} = -1\frac{1}{2}\)

or 

\(\displaystyle x-5= 0\)

\(\displaystyle x=5\)

The solution set is \(\displaystyle \left \{ - 1 \frac{1}{2}, 5\right \}\).

Rewrite this quadratic equation in standard form:

\(\displaystyle x^{2} + 12 = 8x\)

\(\displaystyle x^{2} -8x + 12 = 8x-8x\)

\(\displaystyle x^{2} -8x + 12 = 0\)

Factor the expression on the left. We want two integers whose sum is \(\displaystyle -8\) and whose product is \(\displaystyle 12\). These numbers are \(\displaystyle -6,-2\), so the equation becomes

\(\displaystyle (x-6)(x-2)=0\).

Set each factor equal to 0 separately, then solve:

\(\displaystyle x - 6 = 0 \Rightarrow x = 6\)

\(\displaystyle x - 2 = 0 \Rightarrow x = 2\)

Simplify both sides, then solve:

\(\displaystyle 4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)\)

\(\displaystyle 4 \cdot x + 4 \cdot 6 - 3 \cdot x - 3 \cdot 8 = 2 \cdot x + 2 \cdot 4 +2 \cdot x - 2 \cdot 4\)

\(\displaystyle 4x + 24 - 3 x - 24 = 2x +8 +2 x -8\)

\(\displaystyle 4x - 3 x+ 24 - 24 = 2x +2 x +8-8\)

\(\displaystyle \left ( 4 - 3 \right ) x= \left ( 2 +2 \right ) x\)

\(\displaystyle x= 4 x\)

\(\displaystyle x - 4x = 4 x - 4x\)

\(\displaystyle \left (1 - 4 \right ) x = 0\)

\(\displaystyle -3 x = 0\)

\(\displaystyle x = 0\)

Simplify both sides, then solve:

\(\displaystyle 3 (x + 2) + 4x = 5x + 2 (x +3)\)

\(\displaystyle 3 \cdot x + 3 \cdot 2 + 4x = 5x + 2 \cdot x + 2 \cdot 3\)

\(\displaystyle 3x + 6 + 4x = 5x + 2x + 6\)

\(\displaystyle 3x + 4x + 6 = 5x + 2x + 6\)

\(\displaystyle \left ( 3 + 4 \right ) x + 6 = \left ( 5 + 2 \right ) x + 6\)

\(\displaystyle 7x + 6 =7 x + 6\)

This is an identically true statement, so the original equation has the set of all real numbers as its solution set.

\(\displaystyle 5xp + 3q = W\)

\(\displaystyle 5xp + 3q -3q = W-3q\)

\(\displaystyle 5xp = W-3q\)

\(\displaystyle 5xp \div 5x =\left ( W-3q \right ) \div 5x\)

\(\displaystyle p =\frac{ W-3q }{5x}\)