Solve x^2=-3x+8

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

Solve for x

$$x = \frac{\sqrt{41} - 3}{2} \approx 1.701562119$$

$$x=\frac{-\sqrt{41}-3}{2}\approx -4.701562119$$

Steps Using the Quadratic Formula

Add 3x to both sides.

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

Subtract 8 from both sides.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-3±\sqrt{9+32}}{2}$$

Add 9 to 32.

$$x=\frac{-3±\sqrt{41}}{2}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{2} when ± is plus. Add -3 to \sqrt{41}.

$$x=\frac{\sqrt{41}-3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{2} when ± is minus. Subtract \sqrt{41} from -3.

$$x=\frac{-\sqrt{41}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-3}{2} x=\frac{-\sqrt{41}-3}{2}$$

Steps for Completing the Square

Add 3x to both sides.

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

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.

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

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

$$x^{2}+3x+\frac{9}{4}=8+\frac{9}{4}$$

Add 8 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{41}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{41}}{2} x+\frac{3}{2}=-\frac{\sqrt{41}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{41}-3}{2} x=\frac{-\sqrt{41}-3}{2}$$