Solve x^2-10x=13

$$x ^ { 2 } - 10 x = 13$$

Solve for x

$$x=\sqrt{38}+5\approx 11.164414003$$

$$x=5-\sqrt{38}\approx -1.164414003$$

Steps Using the Quadratic Formula

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.

$$x^{2}-10x=13$$

Subtract 13 from both sides of the equation.

$$x^{2}-10x-13=13-13$$

Subtracting 13 from itself leaves 0.

$$x^{2}-10x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-13\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-13\right)}}{2}$$

Multiply -4 times -13.

$$x=\frac{-\left(-10\right)±\sqrt{100+52}}{2}$$

Add 100 to 52.

$$x=\frac{-\left(-10\right)±\sqrt{152}}{2}$$

Take the square root of 152.

$$x=\frac{-\left(-10\right)±2\sqrt{38}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{38}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{38}}{2} when ± is plus. Add 10 to 2\sqrt{38}.

$$x=\frac{2\sqrt{38}+10}{2}$$

Divide 10+2\sqrt{38} by 2.

$$x=\sqrt{38}+5$$

Now solve the equation x=\frac{10±2\sqrt{38}}{2} when ± is minus. Subtract 2\sqrt{38} from 10.

$$x=\frac{10-2\sqrt{38}}{2}$$

Divide 10-2\sqrt{38} by 2.

$$x=5-\sqrt{38}$$

The equation is now solved.

$$x=\sqrt{38}+5 x=5-\sqrt{38}$$

Steps for Completing the Square

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.

$$x^{2}-10x=13$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=13+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=13+25$$

Add 13 to 25.

$$x^{2}-10x+25=38$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=38$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{38}$$

Simplify.

$$x-5=\sqrt{38} x-5=-\sqrt{38}$$

Add 5 to both sides of the equation.

$$x=\sqrt{38}+5 x=5-\sqrt{38}$$