Solve x+left(x+2right)^2=100

$$x+ { \left(x+2 \right) }^{ 2 } =100$$

Solve for x

$$x = \frac{\sqrt{409} - 5}{2} \approx 7.611874208$$

$$x=\frac{-\sqrt{409}-5}{2}\approx -12.611874208$$

Steps Using the Quadratic Formula

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

$$x+x^{2}+4x+4=100$$

Combine x and 4x to get 5x.

$$5x+x^{2}+4=100$$

Subtract 100 from both sides.

$$5x+x^{2}+4-100=0$$

Subtract 100 from 4 to get -96.

$$5x+x^{2}-96=0$$

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.

$$x^{2}+5x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-96\right)}}{2}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-96\right)}}{2}$$

Multiply -4 times -96.

$$x=\frac{-5±\sqrt{25+384}}{2}$$

Add 25 to 384.

$$x=\frac{-5±\sqrt{409}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{409}}{2} when ± is plus. Add -5 to \sqrt{409}.

$$x=\frac{\sqrt{409}-5}{2}$$

Now solve the equation x=\frac{-5±\sqrt{409}}{2} when ± is minus. Subtract \sqrt{409} from -5.

$$x=\frac{-\sqrt{409}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{409}-5}{2} x=\frac{-\sqrt{409}-5}{2}$$

Steps for Completing the Square

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

$$x+x^{2}+4x+4=100$$

Combine x and 4x to get 5x.

$$5x+x^{2}+4=100$$

Subtract 4 from both sides.

$$5x+x^{2}=100-4$$

Subtract 4 from 100 to get 96.

$$5x+x^{2}=96$$

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.

$$x^{2}+5x=96$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=96+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=96+\frac{25}{4}$$

Add 96 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{409}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{409}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{409}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{409}}{2} x+\frac{5}{2}=-\frac{\sqrt{409}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{409}-5}{2} x=\frac{-\sqrt{409}-5}{2}$$