Solve q^2-36q+540=3q^2-72q+540

$$q ^ { 2 } - 36 q + 540 = 3 q ^ { 2 } - 72 q + 540$$

Solve for q

$$q=18$$

$$q=0$$

Steps Using Factoring

Subtract 3q^{2} from both sides.

$$q^{2}-36q+540-3q^{2}=-72q+540$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

$$-2q^{2}-36q+540=-72q+540$$

Add 72q to both sides.

$$-2q^{2}-36q+540+72q=540$$

Combine -36q and 72q to get 36q.

$$-2q^{2}+36q+540=540$$

Subtract 540 from both sides.

$$-2q^{2}+36q+540-540=0$$

Subtract 540 from 540 to get 0.

$$-2q^{2}+36q=0$$

Factor out q.

$$q\left(-2q+36\right)=0$$

To find equation solutions, solve q=0 and -2q+36=0.

$$q=0 q=18$$

Steps Using the Quadratic Formula

Subtract 3q^{2} from both sides.

$$q^{2}-36q+540-3q^{2}=-72q+540$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

$$-2q^{2}-36q+540=-72q+540$$

Add 72q to both sides.

$$-2q^{2}-36q+540+72q=540$$

Combine -36q and 72q to get 36q.

$$-2q^{2}+36q+540=540$$

Subtract 540 from both sides.

$$-2q^{2}+36q+540-540=0$$

Subtract 540 from 540 to get 0.

$$-2q^{2}+36q=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 36 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-36±\sqrt{36^{2}}}{2\left(-2\right)}$$

Take the square root of 36^{2}.

$$q=\frac{-36±36}{2\left(-2\right)}$$

Multiply 2 times -2.

$$q=\frac{-36±36}{-4}$$

Now solve the equation q=\frac{-36±36}{-4} when ± is plus. Add -36 to 36.

$$q=\frac{0}{-4}$$

Divide 0 by -4.

$$q=0$$

Now solve the equation q=\frac{-36±36}{-4} when ± is minus. Subtract 36 from -36.

$$q=-\frac{72}{-4}$$

Divide -72 by -4.

$$q=18$$

The equation is now solved.

$$q=0 q=18$$

Steps for Completing the Square

Subtract 3q^{2} from both sides.

$$q^{2}-36q+540-3q^{2}=-72q+540$$

Combine q^{2} and -3q^{2} to get -2q^{2}.

$$-2q^{2}-36q+540=-72q+540$$

Add 72q to both sides.

$$-2q^{2}-36q+540+72q=540$$

Combine -36q and 72q to get 36q.

$$-2q^{2}+36q+540=540$$

Subtract 540 from both sides.

$$-2q^{2}+36q=540-540$$

Subtract 540 from 540 to get 0.

$$-2q^{2}+36q=0$$

Divide both sides by -2.

$$\frac{-2q^{2}+36q}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$q^{2}+\frac{36}{-2}q=\frac{0}{-2}$$

Divide 36 by -2.

$$q^{2}-18q=\frac{0}{-2}$$

Divide 0 by -2.

$$q^{2}-18q=0$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-18q+\left(-9\right)^{2}=\left(-9\right)^{2}$$

Square -9.

$$q^{2}-18q+81=81$$

Factor q^{2}-18q+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q-9\right)^{2}=81$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-9\right)^{2}}=\sqrt{81}$$

Simplify.

$$q-9=9 q-9=-9$$

Add 9 to both sides of the equation.

$$q=18 q=0$$