Solve (x+10)(x+12)=3x

$$(x+10)(x+12)=3x$$

Solve for x (complex solution)

$$x=\frac{-19+\sqrt{119}i}{2}\approx -9.5+5.454356057i$$

$$x=\frac{-\sqrt{119}i-19}{2}\approx -9.5-5.454356057i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+10 by x+12 and combine like terms.

$$x^{2}+22x+120=3x$$

Subtract 3x from both sides.

$$x^{2}+22x+120-3x=0$$

Combine 22x and -3x to get 19x.

$$x^{2}+19x+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 19 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-19±\sqrt{19^{2}-4\times 120}}{2}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 120}}{2}$$

Multiply -4 times 120.

$$x=\frac{-19±\sqrt{361-480}}{2}$$

Add 361 to -480.

$$x=\frac{-19±\sqrt{-119}}{2}$$

Take the square root of -119.

$$x=\frac{-19±\sqrt{119}i}{2}$$

Now solve the equation x=\frac{-19±\sqrt{119}i}{2} when ± is plus. Add -19 to i\sqrt{119}.

$$x=\frac{-19+\sqrt{119}i}{2}$$

Now solve the equation x=\frac{-19±\sqrt{119}i}{2} when ± is minus. Subtract i\sqrt{119} from -19.

$$x=\frac{-\sqrt{119}i-19}{2}$$

The equation is now solved.

$$x=\frac{-19+\sqrt{119}i}{2} x=\frac{-\sqrt{119}i-19}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x+10 by x+12 and combine like terms.

$$x^{2}+22x+120=3x$$

Subtract 3x from both sides.

$$x^{2}+22x+120-3x=0$$

Combine 22x and -3x to get 19x.

$$x^{2}+19x+120=0$$

Subtract 120 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}+19x=-120$$

Divide 19, the coefficient of the x term, by 2 to get \frac{19}{2}. Then add the square of \frac{19}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+19x+\left(\frac{19}{2}\right)^{2}=-120+\left(\frac{19}{2}\right)^{2}$$

Square \frac{19}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+19x+\frac{361}{4}=-120+\frac{361}{4}$$

Add -120 to \frac{361}{4}.

$$x^{2}+19x+\frac{361}{4}=-\frac{119}{4}$$

Factor x^{2}+19x+\frac{361}{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}.

$$\left(x+\frac{19}{2}\right)^{2}=-\frac{119}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{2}\right)^{2}}=\sqrt{-\frac{119}{4}}$$

Simplify.

$$x+\frac{19}{2}=\frac{\sqrt{119}i}{2} x+\frac{19}{2}=-\frac{\sqrt{119}i}{2}$$

Subtract \frac{19}{2} from both sides of the equation.

$$x=\frac{-19+\sqrt{119}i}{2} x=\frac{-\sqrt{119}i-19}{2}$$