Solve -9x=13-5x^2

$$- 9 x = 13 - 5 x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{341} + 9}{10} \approx 2.746618531$$

$$x=\frac{9-\sqrt{341}}{10}\approx -0.946618531$$

Steps Using the Quadratic Formula

Subtract 13 from both sides.

$$-9x-13=-5x^{2}$$

Add 5x^{2} to both sides.

$$-9x-13+5x^{2}=0$$

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.

$$5x^{2}-9x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -9 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 5\left(-13\right)}}{2\times 5}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 5\left(-13\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-9\right)±\sqrt{81-20\left(-13\right)}}{2\times 5}$$

Multiply -20 times -13.

$$x=\frac{-\left(-9\right)±\sqrt{81+260}}{2\times 5}$$

Add 81 to 260.

$$x=\frac{-\left(-9\right)±\sqrt{341}}{2\times 5}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{341}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{9±\sqrt{341}}{10}$$

Now solve the equation x=\frac{9±\sqrt{341}}{10} when ± is plus. Add 9 to \sqrt{341}.

$$x=\frac{\sqrt{341}+9}{10}$$

Now solve the equation x=\frac{9±\sqrt{341}}{10} when ± is minus. Subtract \sqrt{341} from 9.

$$x=\frac{9-\sqrt{341}}{10}$$

The equation is now solved.

$$x=\frac{\sqrt{341}+9}{10} x=\frac{9-\sqrt{341}}{10}$$

Steps for Completing the Square

Add 5x^{2} to both sides.

$$-9x+5x^{2}=13$$

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.

$$5x^{2}-9x=13$$

Divide both sides by 5.

$$\frac{5x^{2}-9x}{5}=\frac{13}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{9}{5}x=\frac{13}{5}$$

Divide -\frac{9}{5}, the coefficient of the x term, by 2 to get -\frac{9}{10}. Then add the square of -\frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{5}x+\left(-\frac{9}{10}\right)^{2}=\frac{13}{5}+\left(-\frac{9}{10}\right)^{2}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{13}{5}+\frac{81}{100}$$

Add \frac{13}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{5}x+\frac{81}{100}=\frac{341}{100}$$

Factor x^{2}-\frac{9}{5}x+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{10}\right)^{2}=\frac{341}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{10}\right)^{2}}=\sqrt{\frac{341}{100}}$$

Simplify.

$$x-\frac{9}{10}=\frac{\sqrt{341}}{10} x-\frac{9}{10}=-\frac{\sqrt{341}}{10}$$

Add \frac{9}{10} to both sides of the equation.

$$x=\frac{\sqrt{341}+9}{10} x=\frac{9-\sqrt{341}}{10}$$