Solve x^2=8x+6

$${ x }^{ 2 } =8x+6$$

Solve for x

$$x=\sqrt{22}+4\approx 8.69041576$$

$$x=4-\sqrt{22}\approx -0.69041576$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$x^{2}-8x=6$$

Subtract 6 from both sides.

$$x^{2}-8x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-6\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$x=\frac{-\left(-8\right)±\sqrt{64+24}}{2}$$

Add 64 to 24.

$$x=\frac{-\left(-8\right)±\sqrt{88}}{2}$$

Take the square root of 88.

$$x=\frac{-\left(-8\right)±2\sqrt{22}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{22}}{2}$$

Now solve the equation x=\frac{8±2\sqrt{22}}{2} when ± is plus. Add 8 to 2\sqrt{22}.

$$x=\frac{2\sqrt{22}+8}{2}$$

Divide 8+2\sqrt{22} by 2.

$$x=\sqrt{22}+4$$

Now solve the equation x=\frac{8±2\sqrt{22}}{2} when ± is minus. Subtract 2\sqrt{22} from 8.

$$x=\frac{8-2\sqrt{22}}{2}$$

Divide 8-2\sqrt{22} by 2.

$$x=4-\sqrt{22}$$

The equation is now solved.

$$x=\sqrt{22}+4 x=4-\sqrt{22}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$x^{2}-8x=6$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=6+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=6+16$$

Add 6 to 16.

$$x^{2}-8x+16=22$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=22$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{22}$$

Simplify.

$$x-4=\sqrt{22} x-4=-\sqrt{22}$$

Add 4 to both sides of the equation.

$$x=\sqrt{22}+4 x=4-\sqrt{22}$$