Solve 0=-x^2+3x+2

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

Solve for x

$$x = \frac{\sqrt{17} + 3}{2} \approx 3.561552813$$

$$x=\frac{3-\sqrt{17}}{2}\approx -0.561552813$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-3±\sqrt{9+8}}{2\left(-1\right)}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{17}}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{-2} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{-2}$$

Divide -3+\sqrt{17} by -2.

$$x=\frac{3-\sqrt{17}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{-2}$$

Divide -3-\sqrt{17} by -2.

$$x=\frac{\sqrt{17}+3}{2}$$

The equation is now solved.

$$x=\frac{3-\sqrt{17}}{2} x=\frac{\sqrt{17}+3}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$\frac{-x^{2}+3x}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{3}{-1}x=-\frac{2}{-1}$$

Divide 3 by -1.

$$x^{2}-3x=-\frac{2}{-1}$$

Divide -2 by -1.

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

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=2+\frac{9}{4}$$

Add 2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{17}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{17}}{2} x-\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}$$