Solve x^2+3=8x

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

Solve for x

$$x=\sqrt{13}+4\approx 7.605551275$$

$$x=4-\sqrt{13}\approx 0.394448725$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$x^{2}+3-8x=0$$

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.

$$x^{2}-8x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3}}{2}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12}}{2}$$

Add 64 to -12.

$$x=\frac{-\left(-8\right)±\sqrt{52}}{2}$$

Take the square root of 52.

$$x=\frac{-\left(-8\right)±2\sqrt{13}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{13}}{2}$$

Now solve the equation x=\frac{8±2\sqrt{13}}{2} when ± is plus. Add 8 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+8}{2}$$

Divide 8+2\sqrt{13} by 2.

$$x=\sqrt{13}+4$$

Now solve the equation x=\frac{8±2\sqrt{13}}{2} when ± is minus. Subtract 2\sqrt{13} from 8.

$$x=\frac{8-2\sqrt{13}}{2}$$

Divide 8-2\sqrt{13} by 2.

$$x=4-\sqrt{13}$$

The equation is now solved.

$$x=\sqrt{13}+4 x=4-\sqrt{13}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$x^{2}+3-8x=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

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.

$$x^{2}-8x+\left(-4\right)^{2}=-3+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-3+16$$

Add -3 to 16.

$$x^{2}-8x+16=13$$

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}.

$$\left(x-4\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{13}$$

Simplify.

$$x-4=\sqrt{13} x-4=-\sqrt{13}$$

Add 4 to both sides of the equation.

$$x=\sqrt{13}+4 x=4-\sqrt{13}$$