Solve 2q^2+10q+12=q^2

$$2 q ^ { 2 } + 10 q + 12 = q ^ { 2 }$$

Solve for q (complex solution)

$$q=\sqrt{13}-5\approx -1.394448725$$

$$q=-\left(\sqrt{13}+5\right)\approx -8.605551275$$

Steps Using the Quadratic Formula

Subtract q^{2} from both sides.

$$2q^{2}+10q+12-q^{2}=0$$

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

$$q^{2}+10q+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-10±\sqrt{10^{2}-4\times 12}}{2}$$

Square 10.

$$q=\frac{-10±\sqrt{100-4\times 12}}{2}$$

Multiply -4 times 12.

$$q=\frac{-10±\sqrt{100-48}}{2}$$

Add 100 to -48.

$$q=\frac{-10±\sqrt{52}}{2}$$

Take the square root of 52.

$$q=\frac{-10±2\sqrt{13}}{2}$$

Now solve the equation q=\frac{-10±2\sqrt{13}}{2} when ± is plus. Add -10 to 2\sqrt{13}.

$$q=\frac{2\sqrt{13}-10}{2}$$

Divide -10+2\sqrt{13} by 2.

$$q=\sqrt{13}-5$$

Now solve the equation q=\frac{-10±2\sqrt{13}}{2} when ± is minus. Subtract 2\sqrt{13} from -10.

$$q=\frac{-2\sqrt{13}-10}{2}$$

Divide -10-2\sqrt{13} by 2.

$$q=-\sqrt{13}-5$$

The equation is now solved.

$$q=\sqrt{13}-5 q=-\sqrt{13}-5$$

Steps for Completing the Square

Subtract q^{2} from both sides.

$$2q^{2}+10q+12-q^{2}=0$$

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

$$q^{2}+10q+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$q^{2}+10q=-12$$

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

$$q^{2}+10q+5^{2}=-12+5^{2}$$

Square 5.

$$q^{2}+10q+25=-12+25$$

Add -12 to 25.

$$q^{2}+10q+25=13$$

Factor q^{2}+10q+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+5\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+5\right)^{2}}=\sqrt{13}$$

Simplify.

$$q+5=\sqrt{13} q+5=-\sqrt{13}$$

Subtract 5 from both sides of the equation.

$$q=\sqrt{13}-5 q=-\sqrt{13}-5$$

Solve for q

$$q=\sqrt{13}-5\approx -1.394448725$$

$$q=-\sqrt{13}-5\approx -8.605551275$$

Steps Using the Quadratic Formula

Subtract q^{2} from both sides.

$$2q^{2}+10q+12-q^{2}=0$$

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

$$q^{2}+10q+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-10±\sqrt{10^{2}-4\times 12}}{2}$$

Square 10.

$$q=\frac{-10±\sqrt{100-4\times 12}}{2}$$

Multiply -4 times 12.

$$q=\frac{-10±\sqrt{100-48}}{2}$$

Add 100 to -48.

$$q=\frac{-10±\sqrt{52}}{2}$$

Take the square root of 52.

$$q=\frac{-10±2\sqrt{13}}{2}$$

Now solve the equation q=\frac{-10±2\sqrt{13}}{2} when ± is plus. Add -10 to 2\sqrt{13}.

$$q=\frac{2\sqrt{13}-10}{2}$$

Divide -10+2\sqrt{13} by 2.

$$q=\sqrt{13}-5$$

Now solve the equation q=\frac{-10±2\sqrt{13}}{2} when ± is minus. Subtract 2\sqrt{13} from -10.

$$q=\frac{-2\sqrt{13}-10}{2}$$

Divide -10-2\sqrt{13} by 2.

$$q=-\sqrt{13}-5$$

The equation is now solved.

$$q=\sqrt{13}-5 q=-\sqrt{13}-5$$

Steps for Completing the Square

Subtract q^{2} from both sides.

$$2q^{2}+10q+12-q^{2}=0$$

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

$$q^{2}+10q+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$q^{2}+10q=-12$$

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

$$q^{2}+10q+5^{2}=-12+5^{2}$$

Square 5.

$$q^{2}+10q+25=-12+25$$

Add -12 to 25.

$$q^{2}+10q+25=13$$

Factor q^{2}+10q+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+5\right)^{2}=13$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+5\right)^{2}}=\sqrt{13}$$

Simplify.

$$q+5=\sqrt{13} q+5=-\sqrt{13}$$

Subtract 5 from both sides of the equation.

$$q=\sqrt{13}-5 q=-\sqrt{13}-5$$