Solve 40+40(1+x)+40left(1+xright)^2=0

$$40+40(1+x)+40 { \left(1+x \right) }^{ 2 } =0$$

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{3}i}{2}\approx -1.5+0.866025404i$$

$$x=\frac{-\sqrt{3}i-3}{2}\approx -1.5-0.866025404i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 40 by 1+x.

$$40+40+40x+40\left(1+x\right)^{2}=0$$

Add 40 and 40 to get 80.

$$80+40x+40\left(1+x\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$80+40x+40\left(1+2x+x^{2}\right)=0$$

Use the distributive property to multiply 40 by 1+2x+x^{2}.

$$80+40x+40+80x+40x^{2}=0$$

Add 80 and 40 to get 120.

$$120+40x+80x+40x^{2}=0$$

Combine 40x and 80x to get 120x.

$$120+120x+40x^{2}=0$$

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.

$$40x^{2}+120x+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, 120 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-120±\sqrt{120^{2}-4\times 40\times 120}}{2\times 40}$$

Square 120.

$$x=\frac{-120±\sqrt{14400-4\times 40\times 120}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-120±\sqrt{14400-160\times 120}}{2\times 40}$$

Multiply -160 times 120.

$$x=\frac{-120±\sqrt{14400-19200}}{2\times 40}$$

Add 14400 to -19200.

$$x=\frac{-120±\sqrt{-4800}}{2\times 40}$$

Take the square root of -4800.

$$x=\frac{-120±40\sqrt{3}i}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{-120±40\sqrt{3}i}{80}$$

Now solve the equation x=\frac{-120±40\sqrt{3}i}{80} when ± is plus. Add -120 to 40i\sqrt{3}.

$$x=\frac{-120+40\sqrt{3}i}{80}$$

Divide -120+40i\sqrt{3} by 80.

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

Now solve the equation x=\frac{-120±40\sqrt{3}i}{80} when ± is minus. Subtract 40i\sqrt{3} from -120.

$$x=\frac{-40\sqrt{3}i-120}{80}$$

Divide -120-40i\sqrt{3} by 80.

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 40 by 1+x.

$$40+40+40x+40\left(1+x\right)^{2}=0$$

Add 40 and 40 to get 80.

$$80+40x+40\left(1+x\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$80+40x+40\left(1+2x+x^{2}\right)=0$$

Use the distributive property to multiply 40 by 1+2x+x^{2}.

$$80+40x+40+80x+40x^{2}=0$$

Add 80 and 40 to get 120.

$$120+40x+80x+40x^{2}=0$$

Combine 40x and 80x to get 120x.

$$120+120x+40x^{2}=0$$

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

$$120x+40x^{2}=-120$$

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.

$$40x^{2}+120x=-120$$

Divide both sides by 40.

$$\frac{40x^{2}+120x}{40}=-\frac{120}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\frac{120}{40}x=-\frac{120}{40}$$

Divide 120 by 40.

$$x^{2}+3x=-\frac{120}{40}$$

Divide -120 by 40.

$$x^{2}+3x=-3$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-3+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-3+\frac{9}{4}$$

Add -3 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{3}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

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