Solve -x^2+15x=-2x+10

$$- { x }^{ 2 } +15x=-2x+10$$

Solve for x

$$x = \frac{\sqrt{249} + 17}{2} \approx 16.389866919$$

$$x=\frac{17-\sqrt{249}}{2}\approx 0.610133081$$

Steps Using the Quadratic Formula

Add 2x to both sides.

$$-x^{2}+15x+2x=10$$

Combine 15x and 2x to get 17x.

$$-x^{2}+17x=10$$

Subtract 10 from both sides.

$$-x^{2}+17x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 17 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\left(-1\right)\left(-10\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-17±\sqrt{289+4\left(-10\right)}}{2\left(-1\right)}$$

Multiply 4 times -10.

$$x=\frac{-17±\sqrt{289-40}}{2\left(-1\right)}$$

Add 289 to -40.

$$x=\frac{-17±\sqrt{249}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-17±\sqrt{249}}{-2}$$

Now solve the equation x=\frac{-17±\sqrt{249}}{-2} when ± is plus. Add -17 to \sqrt{249}.

$$x=\frac{\sqrt{249}-17}{-2}$$

Divide -17+\sqrt{249} by -2.

$$x=\frac{17-\sqrt{249}}{2}$$

Now solve the equation x=\frac{-17±\sqrt{249}}{-2} when ± is minus. Subtract \sqrt{249} from -17.

$$x=\frac{-\sqrt{249}-17}{-2}$$

Divide -17-\sqrt{249} by -2.

$$x=\frac{\sqrt{249}+17}{2}$$

The equation is now solved.

$$x=\frac{17-\sqrt{249}}{2} x=\frac{\sqrt{249}+17}{2}$$

Steps for Completing the Square

Add 2x to both sides.

$$-x^{2}+15x+2x=10$$

Combine 15x and 2x to get 17x.

$$-x^{2}+17x=10$$

Divide both sides by -1.

$$\frac{-x^{2}+17x}{-1}=\frac{10}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{17}{-1}x=\frac{10}{-1}$$

Divide 17 by -1.

$$x^{2}-17x=\frac{10}{-1}$$

Divide 10 by -1.

$$x^{2}-17x=-10$$

Divide -17, the coefficient of the x term, by 2 to get -\frac{17}{2}. Then add the square of -\frac{17}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-17x+\left(-\frac{17}{2}\right)^{2}=-10+\left(-\frac{17}{2}\right)^{2}$$

Square -\frac{17}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-17x+\frac{289}{4}=-10+\frac{289}{4}$$

Add -10 to \frac{289}{4}.

$$x^{2}-17x+\frac{289}{4}=\frac{249}{4}$$

Factor x^{2}-17x+\frac{289}{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}.

$$\left(x-\frac{17}{2}\right)^{2}=\frac{249}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{2}\right)^{2}}=\sqrt{\frac{249}{4}}$$

Simplify.

$$x-\frac{17}{2}=\frac{\sqrt{249}}{2} x-\frac{17}{2}=-\frac{\sqrt{249}}{2}$$

Add \frac{17}{2} to both sides of the equation.

$$x=\frac{\sqrt{249}+17}{2} x=\frac{17-\sqrt{249}}{2}$$