Solve 22=39x-5x^2

$$22=39x-5 { x }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{1081} + 39}{10} \approx 7.187856445$$

$$x=\frac{39-\sqrt{1081}}{10}\approx 0.612143555$$

Steps Using the Quadratic Formula

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

$$39x-5x^{2}=22$$

Subtract 22 from both sides.

$$39x-5x^{2}-22=0$$

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.

$$-5x^{2}+39x-22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 39 for b, and -22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-39±\sqrt{39^{2}-4\left(-5\right)\left(-22\right)}}{2\left(-5\right)}$$

Square 39.

$$x=\frac{-39±\sqrt{1521-4\left(-5\right)\left(-22\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-39±\sqrt{1521+20\left(-22\right)}}{2\left(-5\right)}$$

Multiply 20 times -22.

$$x=\frac{-39±\sqrt{1521-440}}{2\left(-5\right)}$$

Add 1521 to -440.

$$x=\frac{-39±\sqrt{1081}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-39±\sqrt{1081}}{-10}$$

Now solve the equation x=\frac{-39±\sqrt{1081}}{-10} when ± is plus. Add -39 to \sqrt{1081}.

$$x=\frac{\sqrt{1081}-39}{-10}$$

Divide -39+\sqrt{1081} by -10.

$$x=\frac{39-\sqrt{1081}}{10}$$

Now solve the equation x=\frac{-39±\sqrt{1081}}{-10} when ± is minus. Subtract \sqrt{1081} from -39.

$$x=\frac{-\sqrt{1081}-39}{-10}$$

Divide -39-\sqrt{1081} by -10.

$$x=\frac{\sqrt{1081}+39}{10}$$

The equation is now solved.

$$x=\frac{39-\sqrt{1081}}{10} x=\frac{\sqrt{1081}+39}{10}$$

Steps for Completing the Square

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

$$39x-5x^{2}=22$$

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.

$$-5x^{2}+39x=22$$

Divide both sides by -5.

$$\frac{-5x^{2}+39x}{-5}=\frac{22}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{39}{-5}x=\frac{22}{-5}$$

Divide 39 by -5.

$$x^{2}-\frac{39}{5}x=\frac{22}{-5}$$

Divide 22 by -5.

$$x^{2}-\frac{39}{5}x=-\frac{22}{5}$$

Divide -\frac{39}{5}, the coefficient of the x term, by 2 to get -\frac{39}{10}. Then add the square of -\frac{39}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{39}{5}x+\left(-\frac{39}{10}\right)^{2}=-\frac{22}{5}+\left(-\frac{39}{10}\right)^{2}$$

Square -\frac{39}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{39}{5}x+\frac{1521}{100}=-\frac{22}{5}+\frac{1521}{100}$$

Add -\frac{22}{5} to \frac{1521}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{39}{5}x+\frac{1521}{100}=\frac{1081}{100}$$

Factor x^{2}-\frac{39}{5}x+\frac{1521}{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}.

$$\left(x-\frac{39}{10}\right)^{2}=\frac{1081}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{39}{10}\right)^{2}}=\sqrt{\frac{1081}{100}}$$

Simplify.

$$x-\frac{39}{10}=\frac{\sqrt{1081}}{10} x-\frac{39}{10}=-\frac{\sqrt{1081}}{10}$$

Add \frac{39}{10} to both sides of the equation.

$$x=\frac{\sqrt{1081}+39}{10} x=\frac{39-\sqrt{1081}}{10}$$