Solve -3x^2-4x-88=-4x^2+8

$$- 3 x ^ { 2 } - 4 x - 88 = - 4 x ^ { 2 } + 8$$

Solve for x

$$x=-8$$

$$x=12$$

Steps Using Factoring

Add 4x^{2} to both sides.

$$-3x^{2}-4x-88+4x^{2}=8$$

Combine -3x^{2} and 4x^{2} to get x^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from -88 to get -96.

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

To solve the equation, factor x^{2}-4x-96 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=-96$$

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 -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -4.

$$a=-12 b=8$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-12\right)\left(x+8\right)$$

To find equation solutions, solve x-12=0 and x+8=0.

$$x=12 x=-8$$

Steps Using Factoring By Grouping

Add 4x^{2} to both sides.

$$-3x^{2}-4x-88+4x^{2}=8$$

Combine -3x^{2} and 4x^{2} to get x^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from -88 to get -96.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-96\right)=-96$$

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 -96.

$$1,-96 2,-48 3,-32 4,-24 6,-16 8,-12$$

Calculate the sum for each pair.

$$1-96=-95 2-48=-46 3-32=-29 4-24=-20 6-16=-10 8-12=-4$$

The solution is the pair that gives sum -4.

$$a=-12 b=8$$

Rewrite x^{2}-4x-96 as \left(x^{2}-12x\right)+\left(8x-96\right).

$$\left(x^{2}-12x\right)+\left(8x-96\right)$$

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

$$x\left(x-12\right)+8\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+8\right)$$

To find equation solutions, solve x-12=0 and x+8=0.

$$x=12 x=-8$$

Steps Using the Quadratic Formula

Add 4x^{2} to both sides.

$$-3x^{2}-4x-88+4x^{2}=8$$

Combine -3x^{2} and 4x^{2} to get x^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from -88 to get -96.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

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

Multiply -4 times -96.

$$x=\frac{-\left(-4\right)±\sqrt{16+384}}{2}$$

Add 16 to 384.

$$x=\frac{-\left(-4\right)±\sqrt{400}}{2}$$

Take the square root of 400.

$$x=\frac{-\left(-4\right)±20}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±20}{2}$$

Now solve the equation x=\frac{4±20}{2} when ± is plus. Add 4 to 20.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{4±20}{2} when ± is minus. Subtract 20 from 4.

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

The equation is now solved.

$$x=12 x=-8$$

Steps for Completing the Square

Add 4x^{2} to both sides.

$$-3x^{2}-4x-88+4x^{2}=8$$

Combine -3x^{2} and 4x^{2} to get x^{2}.

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

Add 88 to both sides.

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

Add 8 and 88 to get 96.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 96 to 4.

$$x^{2}-4x+4=100$$

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{100}$$

Simplify.

$$x-2=10 x-2=-10$$

Add 2 to both sides of the equation.

$$x=12 x=-8$$