Solve 50(1-10%)(1+x)^2=668

$$50 ( 1 - 10 \% ) ( 1 + x ) ^ { 2 } = 668$$

Solve for x

$$x=\frac{2\sqrt{835}}{15}-1\approx 2.852848874$$

$$x=-\frac{2\sqrt{835}}{15}-1\approx -4.852848874$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{10}{100} to lowest terms by extracting and canceling out 10.

$$50\left(1-\frac{1}{10}\right)\left(1+x\right)^{2}=668$$

Subtract \frac{1}{10} from 1 to get \frac{9}{10}.

$$50\times \frac{9}{10}\left(1+x\right)^{2}=668$$

Multiply 50 and \frac{9}{10} to get 45.

$$45\left(1+x\right)^{2}=668$$

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

$$45\left(1+2x+x^{2}\right)=668$$

Use the distributive property to multiply 45 by 1+2x+x^{2}.

$$45+90x+45x^{2}=668$$

Subtract 668 from both sides.

$$45+90x+45x^{2}-668=0$$

Subtract 668 from 45 to get -623.

$$-623+90x+45x^{2}=0$$

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.

$$45x^{2}+90x-623=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 45 for a, 90 for b, and -623 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-90±\sqrt{90^{2}-4\times 45\left(-623\right)}}{2\times 45}$$

Square 90.

$$x=\frac{-90±\sqrt{8100-4\times 45\left(-623\right)}}{2\times 45}$$

Multiply -4 times 45.

$$x=\frac{-90±\sqrt{8100-180\left(-623\right)}}{2\times 45}$$

Multiply -180 times -623.

$$x=\frac{-90±\sqrt{8100+112140}}{2\times 45}$$

Add 8100 to 112140.

$$x=\frac{-90±\sqrt{120240}}{2\times 45}$$

Take the square root of 120240.

$$x=\frac{-90±12\sqrt{835}}{2\times 45}$$

Multiply 2 times 45.

$$x=\frac{-90±12\sqrt{835}}{90}$$

Now solve the equation x=\frac{-90±12\sqrt{835}}{90} when ± is plus. Add -90 to 12\sqrt{835}.

$$x=\frac{12\sqrt{835}-90}{90}$$

Divide -90+12\sqrt{835} by 90.

$$x=\frac{2\sqrt{835}}{15}-1$$

Now solve the equation x=\frac{-90±12\sqrt{835}}{90} when ± is minus. Subtract 12\sqrt{835} from -90.

$$x=\frac{-12\sqrt{835}-90}{90}$$

Divide -90-12\sqrt{835} by 90.

$$x=-\frac{2\sqrt{835}}{15}-1$$

The equation is now solved.

$$x=\frac{2\sqrt{835}}{15}-1 x=-\frac{2\sqrt{835}}{15}-1$$

Steps for Completing the Square

Reduce the fraction \frac{10}{100} to lowest terms by extracting and canceling out 10.

$$50\left(1-\frac{1}{10}\right)\left(1+x\right)^{2}=668$$

Subtract \frac{1}{10} from 1 to get \frac{9}{10}.

$$50\times \frac{9}{10}\left(1+x\right)^{2}=668$$

Multiply 50 and \frac{9}{10} to get 45.

$$45\left(1+x\right)^{2}=668$$

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

$$45\left(1+2x+x^{2}\right)=668$$

Use the distributive property to multiply 45 by 1+2x+x^{2}.

$$45+90x+45x^{2}=668$$

Subtract 45 from both sides.

$$90x+45x^{2}=668-45$$

Subtract 45 from 668 to get 623.

$$90x+45x^{2}=623$$

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.

$$45x^{2}+90x=623$$

Divide both sides by 45.

$$\frac{45x^{2}+90x}{45}=\frac{623}{45}$$

Dividing by 45 undoes the multiplication by 45.

$$x^{2}+\frac{90}{45}x=\frac{623}{45}$$

Divide 90 by 45.

$$x^{2}+2x=\frac{623}{45}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{623}{45}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{623}{45}+1$$

Add \frac{623}{45} to 1.

$$x^{2}+2x+1=\frac{668}{45}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{668}{45}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{668}{45}}$$

Simplify.

$$x+1=\frac{2\sqrt{835}}{15} x+1=-\frac{2\sqrt{835}}{15}$$

Subtract 1 from both sides of the equation.

$$x=\frac{2\sqrt{835}}{15}-1 x=-\frac{2\sqrt{835}}{15}-1$$