Solve 3x(-8+x)=x^2+8

$$3 x ( - 8 + x ) = x ^ { 2 } + 8$$

Solve for x

$$x=2\sqrt{10}+6\approx 12.32455532$$

$$x=6-2\sqrt{10}\approx -0.32455532$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by -8+x.

$$-24x+3x^{2}=x^{2}+8$$

Subtract x^{2} from both sides.

$$-24x+3x^{2}-x^{2}=8$$

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

$$-24x+2x^{2}=8$$

Subtract 8 from both sides.

$$-24x+2x^{2}-8=0$$

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.

$$2x^{2}-24x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -24 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 2\left(-8\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-24\right)±\sqrt{576-8\left(-8\right)}}{2\times 2}$$

Multiply -8 times -8.

$$x=\frac{-\left(-24\right)±\sqrt{576+64}}{2\times 2}$$

Add 576 to 64.

$$x=\frac{-\left(-24\right)±\sqrt{640}}{2\times 2}$$

Take the square root of 640.

$$x=\frac{-\left(-24\right)±8\sqrt{10}}{2\times 2}$$

The opposite of -24 is 24.

$$x=\frac{24±8\sqrt{10}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{24±8\sqrt{10}}{4}$$

Now solve the equation x=\frac{24±8\sqrt{10}}{4} when ± is plus. Add 24 to 8\sqrt{10}.

$$x=\frac{8\sqrt{10}+24}{4}$$

Divide 24+8\sqrt{10} by 4.

$$x=2\sqrt{10}+6$$

Now solve the equation x=\frac{24±8\sqrt{10}}{4} when ± is minus. Subtract 8\sqrt{10} from 24.

$$x=\frac{24-8\sqrt{10}}{4}$$

Divide 24-8\sqrt{10} by 4.

$$x=6-2\sqrt{10}$$

The equation is now solved.

$$x=2\sqrt{10}+6 x=6-2\sqrt{10}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by -8+x.

$$-24x+3x^{2}=x^{2}+8$$

Subtract x^{2} from both sides.

$$-24x+3x^{2}-x^{2}=8$$

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

$$-24x+2x^{2}=8$$

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.

$$2x^{2}-24x=8$$

Divide both sides by 2.

$$\frac{2x^{2}-24x}{2}=\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{24}{2}\right)x=\frac{8}{2}$$

Divide -24 by 2.

$$x^{2}-12x=\frac{8}{2}$$

Divide 8 by 2.

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

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=4+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=4+36$$

Add 4 to 36.

$$x^{2}-12x+36=40$$

Factor x^{2}-12x+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}.

$$\left(x-6\right)^{2}=40$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{40}$$

Simplify.

$$x-6=2\sqrt{10} x-6=-2\sqrt{10}$$

Add 6 to both sides of the equation.

$$x=2\sqrt{10}+6 x=6-2\sqrt{10}$$