Solve 9(30-x)=x^2

$$9(30-x)= { x }^{ 2 }$$

Solve for x

$$x = \frac{3 \sqrt{129} - 9}{2} \approx 12.536725037$$

$$x=\frac{-3\sqrt{129}-9}{2}\approx -21.536725037$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9 by 30-x.

$$270-9x=x^{2}$$

Subtract x^{2} from both sides.

$$270-9x-x^{2}=0$$

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.

$$-x^{2}-9x+270=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -9 for b, and 270 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-1\right)\times 270}}{2\left(-1\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-1\right)\times 270}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+4\times 270}}{2\left(-1\right)}$$

Multiply 4 times 270.

$$x=\frac{-\left(-9\right)±\sqrt{81+1080}}{2\left(-1\right)}$$

Add 81 to 1080.

$$x=\frac{-\left(-9\right)±\sqrt{1161}}{2\left(-1\right)}$$

Take the square root of 1161.

$$x=\frac{-\left(-9\right)±3\sqrt{129}}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±3\sqrt{129}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{9±3\sqrt{129}}{-2}$$

Now solve the equation x=\frac{9±3\sqrt{129}}{-2} when ± is plus. Add 9 to 3\sqrt{129}.

$$x=\frac{3\sqrt{129}+9}{-2}$$

Divide 9+3\sqrt{129} by -2.

$$x=\frac{-3\sqrt{129}-9}{2}$$

Now solve the equation x=\frac{9±3\sqrt{129}}{-2} when ± is minus. Subtract 3\sqrt{129} from 9.

$$x=\frac{9-3\sqrt{129}}{-2}$$

Divide 9-3\sqrt{129} by -2.

$$x=\frac{3\sqrt{129}-9}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{129}-9}{2} x=\frac{3\sqrt{129}-9}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 9 by 30-x.

$$270-9x=x^{2}$$

Subtract x^{2} from both sides.

$$270-9x-x^{2}=0$$

Subtract 270 from both sides. Anything subtracted from zero gives its negation.

$$-9x-x^{2}=-270$$

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.

$$-x^{2}-9x=-270$$

Divide both sides by -1.

$$\frac{-x^{2}-9x}{-1}=-\frac{270}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{9}{-1}\right)x=-\frac{270}{-1}$$

Divide -9 by -1.

$$x^{2}+9x=-\frac{270}{-1}$$

Divide -270 by -1.

$$x^{2}+9x=270$$

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

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=270+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=270+\frac{81}{4}$$

Add 270 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{1161}{4}$$

Factor x^{2}+9x+\frac{81}{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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{1161}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{1161}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{3\sqrt{129}}{2} x+\frac{9}{2}=-\frac{3\sqrt{129}}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=\frac{3\sqrt{129}-9}{2} x=\frac{-3\sqrt{129}-9}{2}$$