Solve -270x=30x^2

$$- 270 x = 30 x ^ { 2 }$$

Solve for x

$$x=-9$$

$$x=0$$

Steps Using Factoring

Subtract 30x^{2} from both sides.

$$-270x-30x^{2}=0$$

Factor out x.

$$x\left(-270-30x\right)=0$$

To find equation solutions, solve x=0 and -270-30x=0.

$$x=0 x=-9$$

Steps Using the Quadratic Formula

Subtract 30x^{2} from both sides.

$$-270x-30x^{2}=0$$

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.

$$-30x^{2}-270x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -30 for a, -270 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-270\right)±\sqrt{\left(-270\right)^{2}}}{2\left(-30\right)}$$

Take the square root of \left(-270\right)^{2}.

$$x=\frac{-\left(-270\right)±270}{2\left(-30\right)}$$

The opposite of -270 is 270.

$$x=\frac{270±270}{2\left(-30\right)}$$

Multiply 2 times -30.

$$x=\frac{270±270}{-60}$$

Now solve the equation x=\frac{270±270}{-60} when ± is plus. Add 270 to 270.

$$x=\frac{540}{-60}$$

Divide 540 by -60.

$$x=-9$$

Now solve the equation x=\frac{270±270}{-60} when ± is minus. Subtract 270 from 270.

$$x=\frac{0}{-60}$$

Divide 0 by -60.

$$x=0$$

The equation is now solved.

$$x=-9 x=0$$

Steps for Completing the Square

Subtract 30x^{2} from both sides.

$$-270x-30x^{2}=0$$

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.

$$-30x^{2}-270x=0$$

Divide both sides by -30.

$$\frac{-30x^{2}-270x}{-30}=\frac{0}{-30}$$

Dividing by -30 undoes the multiplication by -30.

$$x^{2}+\left(-\frac{270}{-30}\right)x=\frac{0}{-30}$$

Divide -270 by -30.

$$x^{2}+9x=\frac{0}{-30}$$

Divide 0 by -30.

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

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.

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

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

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

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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{9}{2} x+\frac{9}{2}=-\frac{9}{2}$$

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

$$x=0 x=-9$$