Solve 60left(1+xright)^2+60left(1+xright)+60=198.6

$$60 { \left(1+x \right) }^{ 2 } +60 \left( 1+x \right) +60=198.6$$

Solve for x

$$x=0.1$$

$$x=-3.1$$

Steps Using the Quadratic Formula

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

$$60\left(1+2x+x^{2}\right)+60\left(1+x\right)+60=198.6$$

Use the distributive property to multiply 60 by 1+2x+x^{2}.

$$60+120x+60x^{2}+60\left(1+x\right)+60=198.6$$

Use the distributive property to multiply 60 by 1+x.

$$60+120x+60x^{2}+60+60x+60=198.6$$

Add 60 and 60 to get 120.

$$120+120x+60x^{2}+60x+60=198.6$$

Combine 120x and 60x to get 180x.

$$120+180x+60x^{2}+60=198.6$$

Add 120 and 60 to get 180.

$$180+180x+60x^{2}=198.6$$

Subtract 198.6 from both sides.

$$180+180x+60x^{2}-198.6=0$$

Subtract 198.6 from 180 to get -18.6.

$$-18.6+180x+60x^{2}=0$$

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.

$$60x^{2}+180x-18.6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 60 for a, 180 for b, and -18.6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-180±\sqrt{180^{2}-4\times 60\left(-18.6\right)}}{2\times 60}$$

Square 180.

$$x=\frac{-180±\sqrt{32400-4\times 60\left(-18.6\right)}}{2\times 60}$$

Multiply -4 times 60.

$$x=\frac{-180±\sqrt{32400-240\left(-18.6\right)}}{2\times 60}$$

Multiply -240 times -18.6.

$$x=\frac{-180±\sqrt{32400+4464}}{2\times 60}$$

Add 32400 to 4464.

$$x=\frac{-180±\sqrt{36864}}{2\times 60}$$

Take the square root of 36864.

$$x=\frac{-180±192}{2\times 60}$$

Multiply 2 times 60.

$$x=\frac{-180±192}{120}$$

Now solve the equation x=\frac{-180±192}{120} when ± is plus. Add -180 to 192.

$$x=\frac{12}{120}$$

Reduce the fraction \frac{12}{120} to lowest terms by extracting and canceling out 12.

$$x=\frac{1}{10}$$

Now solve the equation x=\frac{-180±192}{120} when ± is minus. Subtract 192 from -180.

$$x=-\frac{372}{120}$$

Reduce the fraction \frac{-372}{120} to lowest terms by extracting and canceling out 12.

$$x=-\frac{31}{10}$$

The equation is now solved.

$$x=\frac{1}{10} x=-\frac{31}{10}$$

Steps for Completing the Square

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

$$60\left(1+2x+x^{2}\right)+60\left(1+x\right)+60=198.6$$

Use the distributive property to multiply 60 by 1+2x+x^{2}.

$$60+120x+60x^{2}+60\left(1+x\right)+60=198.6$$

Use the distributive property to multiply 60 by 1+x.

$$60+120x+60x^{2}+60+60x+60=198.6$$

Add 60 and 60 to get 120.

$$120+120x+60x^{2}+60x+60=198.6$$

Combine 120x and 60x to get 180x.

$$120+180x+60x^{2}+60=198.6$$

Add 120 and 60 to get 180.

$$180+180x+60x^{2}=198.6$$

Subtract 180 from both sides.

$$180x+60x^{2}=198.6-180$$

Subtract 180 from 198.6 to get 18.6.

$$180x+60x^{2}=18.6$$

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.

$$60x^{2}+180x=18.6$$

Divide both sides by 60.

$$\frac{60x^{2}+180x}{60}=\frac{18.6}{60}$$

Dividing by 60 undoes the multiplication by 60.

$$x^{2}+\frac{180}{60}x=\frac{18.6}{60}$$

Divide 180 by 60.

$$x^{2}+3x=\frac{18.6}{60}$$

Divide 18.6 by 60.

$$x^{2}+3x=0.31$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=0.31+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=0.31+\frac{9}{4}$$

Add 0.31 to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{64}{25}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{64}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{64}{25}}$$

Simplify.

$$x+\frac{3}{2}=\frac{8}{5} x+\frac{3}{2}=-\frac{8}{5}$$

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

$$x=\frac{1}{10} x=-\frac{31}{10}$$