Solve 12x-240=1/5x^2

$$12x-240= \frac{ 1 }{ 5 } { x }^{ 2 }$$

Solve for x (complex solution)

$$x=30+10\sqrt{3}i\approx 30+17.320508076i$$

$$x=-10\sqrt{3}i+30\approx 30-17.320508076i$$

Steps Using the Quadratic Formula

Subtract \frac{1}{5}x^{2} from both sides.

$$12x-240-\frac{1}{5}x^{2}=0$$

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.

$$-\frac{1}{5}x^{2}+12x-240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{5} for a, 12 for b, and -240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-\frac{1}{5}\right)\left(-240\right)}}{2\left(-\frac{1}{5}\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-\frac{1}{5}\right)\left(-240\right)}}{2\left(-\frac{1}{5}\right)}$$

Multiply -4 times -\frac{1}{5}.

$$x=\frac{-12±\sqrt{144+\frac{4}{5}\left(-240\right)}}{2\left(-\frac{1}{5}\right)}$$

Multiply \frac{4}{5} times -240.

$$x=\frac{-12±\sqrt{144-192}}{2\left(-\frac{1}{5}\right)}$$

Add 144 to -192.

$$x=\frac{-12±\sqrt{-48}}{2\left(-\frac{1}{5}\right)}$$

Take the square root of -48.

$$x=\frac{-12±4\sqrt{3}i}{2\left(-\frac{1}{5}\right)}$$

Multiply 2 times -\frac{1}{5}.

$$x=\frac{-12±4\sqrt{3}i}{-\frac{2}{5}}$$

Now solve the equation x=\frac{-12±4\sqrt{3}i}{-\frac{2}{5}} when ± is plus. Add -12 to 4i\sqrt{3}.

$$x=\frac{-12+4\sqrt{3}i}{-\frac{2}{5}}$$

Divide -12+4i\sqrt{3} by -\frac{2}{5} by multiplying -12+4i\sqrt{3} by the reciprocal of -\frac{2}{5}.

$$x=-10\sqrt{3}i+30$$

Now solve the equation x=\frac{-12±4\sqrt{3}i}{-\frac{2}{5}} when ± is minus. Subtract 4i\sqrt{3} from -12.

$$x=\frac{-4\sqrt{3}i-12}{-\frac{2}{5}}$$

Divide -12-4i\sqrt{3} by -\frac{2}{5} by multiplying -12-4i\sqrt{3} by the reciprocal of -\frac{2}{5}.

$$x=30+10\sqrt{3}i$$

The equation is now solved.

$$x=-10\sqrt{3}i+30 x=30+10\sqrt{3}i$$

Steps for Completing the Square

Subtract \frac{1}{5}x^{2} from both sides.

$$12x-240-\frac{1}{5}x^{2}=0$$

Add 240 to both sides. Anything plus zero gives itself.

$$12x-\frac{1}{5}x^{2}=240$$

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.

$$-\frac{1}{5}x^{2}+12x=240$$

Multiply both sides by -5.

$$\frac{-\frac{1}{5}x^{2}+12x}{-\frac{1}{5}}=\frac{240}{-\frac{1}{5}}$$

Dividing by -\frac{1}{5} undoes the multiplication by -\frac{1}{5}.

$$x^{2}+\frac{12}{-\frac{1}{5}}x=\frac{240}{-\frac{1}{5}}$$

Divide 12 by -\frac{1}{5} by multiplying 12 by the reciprocal of -\frac{1}{5}.

$$x^{2}-60x=\frac{240}{-\frac{1}{5}}$$

Divide 240 by -\frac{1}{5} by multiplying 240 by the reciprocal of -\frac{1}{5}.

$$x^{2}-60x=-1200$$

Divide -60, the coefficient of the x term, by 2 to get -30. Then add the square of -30 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-60x+\left(-30\right)^{2}=-1200+\left(-30\right)^{2}$$

Square -30.

$$x^{2}-60x+900=-1200+900$$

Add -1200 to 900.

$$x^{2}-60x+900=-300$$

Factor x^{2}-60x+900. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-30\right)^{2}=-300$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-30\right)^{2}}=\sqrt{-300}$$

Simplify.

$$x-30=10\sqrt{3}i x-30=-10\sqrt{3}i$$

Add 30 to both sides of the equation.

$$x=30+10\sqrt{3}i x=-10\sqrt{3}i+30$$