Solve (2x-8)*x=23

$$( 2 x - 8 ) \cdot x = 23$$

Solve for x

$$x=\frac{\sqrt{62}}{2}+2\approx 5.937003937$$

$$x=-\frac{\sqrt{62}}{2}+2\approx -1.937003937$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-8 by x.

$$2x^{2}-8x=23$$

Subtract 23 from both sides.

$$2x^{2}-8x-23=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-23\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-23\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-23\right)}}{2\times 2}$$

Multiply -8 times -23.

$$x=\frac{-\left(-8\right)±\sqrt{64+184}}{2\times 2}$$

Add 64 to 184.

$$x=\frac{-\left(-8\right)±\sqrt{248}}{2\times 2}$$

Take the square root of 248.

$$x=\frac{-\left(-8\right)±2\sqrt{62}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{62}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±2\sqrt{62}}{4}$$

Now solve the equation x=\frac{8±2\sqrt{62}}{4} when ± is plus. Add 8 to 2\sqrt{62}.

$$x=\frac{2\sqrt{62}+8}{4}$$

Divide 8+2\sqrt{62} by 4.

$$x=\frac{\sqrt{62}}{2}+2$$

Now solve the equation x=\frac{8±2\sqrt{62}}{4} when ± is minus. Subtract 2\sqrt{62} from 8.

$$x=\frac{8-2\sqrt{62}}{4}$$

Divide 8-2\sqrt{62} by 4.

$$x=-\frac{\sqrt{62}}{2}+2$$

The equation is now solved.

$$x=\frac{\sqrt{62}}{2}+2 x=-\frac{\sqrt{62}}{2}+2$$

Steps for Completing the Square

Use the distributive property to multiply 2x-8 by x.

$$2x^{2}-8x=23$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{23}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{23}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{23}{2}$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=\frac{23}{2}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=\frac{23}{2}+4$$

Add \frac{23}{2} to 4.

$$x^{2}-4x+4=\frac{31}{2}$$

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}.

$$\left(x-2\right)^{2}=\frac{31}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{31}{2}}$$

Simplify.

$$x-2=\frac{\sqrt{62}}{2} x-2=-\frac{\sqrt{62}}{2}$$

Add 2 to both sides of the equation.

$$x=\frac{\sqrt{62}}{2}+2 x=-\frac{\sqrt{62}}{2}+2$$