Solve 8x^2-4x-88=-4x^2+8

$$8 x ^ { 2 } - 4 x - 88 = - 4 x ^ { 2 } + 8$$

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$x=3$$

Steps Using Factoring By Grouping

Add 4x^{2} to both sides.

$$8x^{2}-4x-88+4x^{2}=8$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}-4x-88=8$$

Subtract 8 from both sides.

$$12x^{2}-4x-88-8=0$$

Subtract 8 from -88 to get -96.

$$12x^{2}-4x-96=0$$

Divide both sides by 4.

$$3x^{2}-x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-24\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=-9 b=8$$

Rewrite 3x^{2}-x-24 as \left(3x^{2}-9x\right)+\left(8x-24\right).

$$\left(3x^{2}-9x\right)+\left(8x-24\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(x-3\right)+8\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(3x+8\right)$$

To find equation solutions, solve x-3=0 and 3x+8=0.

$$x=3 x=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Add 4x^{2} to both sides.

$$8x^{2}-4x-88+4x^{2}=8$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}-4x-88=8$$

Subtract 8 from both sides.

$$12x^{2}-4x-88-8=0$$

Subtract 8 from -88 to get -96.

$$12x^{2}-4x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -4 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 12\left(-96\right)}}{2\times 12}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 12\left(-96\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-4\right)±\sqrt{16-48\left(-96\right)}}{2\times 12}$$

Multiply -48 times -96.

$$x=\frac{-\left(-4\right)±\sqrt{16+4608}}{2\times 12}$$

Add 16 to 4608.

$$x=\frac{-\left(-4\right)±\sqrt{4624}}{2\times 12}$$

Take the square root of 4624.

$$x=\frac{-\left(-4\right)±68}{2\times 12}$$

The opposite of -4 is 4.

$$x=\frac{4±68}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{4±68}{24}$$

Now solve the equation x=\frac{4±68}{24} when ± is plus. Add 4 to 68.

$$x=\frac{72}{24}$$

Divide 72 by 24.

$$x=3$$

Now solve the equation x=\frac{4±68}{24} when ± is minus. Subtract 68 from 4.

$$x=-\frac{64}{24}$$

Reduce the fraction \frac{-64}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{8}{3}$$

The equation is now solved.

$$x=3 x=-\frac{8}{3}$$

Steps for Completing the Square

Add 4x^{2} to both sides.

$$8x^{2}-4x-88+4x^{2}=8$$

Combine 8x^{2} and 4x^{2} to get 12x^{2}.

$$12x^{2}-4x-88=8$$

Add 88 to both sides.

$$12x^{2}-4x=8+88$$

Add 8 and 88 to get 96.

$$12x^{2}-4x=96$$

Divide both sides by 12.

$$\frac{12x^{2}-4x}{12}=\frac{96}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{4}{12}\right)x=\frac{96}{12}$$

Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{1}{3}x=\frac{96}{12}$$

Divide 96 by 12.

$$x^{2}-\frac{1}{3}x=8$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=8+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=8+\frac{1}{36}$$

Add 8 to \frac{1}{36}.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{289}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{289}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{289}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{17}{6} x-\frac{1}{6}=-\frac{17}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=3 x=-\frac{8}{3}$$