Solve 4x-8=4x^2-12x-8

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

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Subtract 4x^{2} from both sides.

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

Add 12x to both sides.

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

Combine 4x and 12x to get 16x.

$$16x-8-4x^{2}=-8$$

Add 8 to both sides.

$$16x-8-4x^{2}+8=0$$

Add -8 and 8 to get 0.

$$16x-4x^{2}=0$$

Factor out x.

$$x\left(16-4x\right)=0$$

To find equation solutions, solve x=0 and 16-4x=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

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

Add 12x to both sides.

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

Combine 4x and 12x to get 16x.

$$16x-8-4x^{2}=-8$$

Add 8 to both sides.

$$16x-8-4x^{2}+8=0$$

Add -8 and 8 to get 0.

$$16x-4x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4x^{2}+16x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 16^{2}.

$$x=\frac{-16±16}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-16±16}{-8}$$

Now solve the equation x=\frac{-16±16}{-8} when ± is plus. Add -16 to 16.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-16±16}{-8} when ± is minus. Subtract 16 from -16.

$$x=-\frac{32}{-8}$$

Divide -32 by -8.

$$x=4$$

The equation is now solved.

$$x=0 x=4$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

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

Add 12x to both sides.

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

Combine 4x and 12x to get 16x.

$$16x-8-4x^{2}=-8$$

Add 8 to both sides.

$$16x-4x^{2}=-8+8$$

Add -8 and 8 to get 0.

$$16x-4x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+16x=0$$

Divide both sides by -4.

$$\frac{-4x^{2}+16x}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{16}{-4}x=\frac{0}{-4}$$

Divide 16 by -4.

$$x^{2}-4x=\frac{0}{-4}$$

Divide 0 by -4.

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

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.

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

Square -2.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

$$x=4 x=0$$