Solve 6^2=(5+x)^2+(x+2)^2

$$6 ^ { 2 } = {( 5 + x )} ^ { 2 } + {( x + 2 )} ^ { 2 }$$

Solve for x

$$x=\frac{3\sqrt{7}-7}{2}\approx 0.468626967$$

$$x=\frac{-3\sqrt{7}-7}{2}\approx -7.468626967$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 2 and get 36.

$$36=\left(5+x\right)^{2}+\left(x+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+x\right)^{2}.

$$36=25+10x+x^{2}+\left(x+2\right)^{2}$$

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

$$36=25+10x+x^{2}+x^{2}+4x+4$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$36=25+10x+2x^{2}+4x+4$$

Combine 10x and 4x to get 14x.

$$36=25+14x+2x^{2}+4$$

Add 25 and 4 to get 29.

$$36=29+14x+2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$29+14x+2x^{2}=36$$

Subtract 36 from both sides.

$$29+14x+2x^{2}-36=0$$

Subtract 36 from 29 to get -7.

$$-7+14x+2x^{2}=0$$

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.

$$2x^{2}+14x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 14 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 2\left(-7\right)}}{2\times 2}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 2\left(-7\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-14±\sqrt{196-8\left(-7\right)}}{2\times 2}$$

Multiply -8 times -7.

$$x=\frac{-14±\sqrt{196+56}}{2\times 2}$$

Add 196 to 56.

$$x=\frac{-14±\sqrt{252}}{2\times 2}$$

Take the square root of 252.

$$x=\frac{-14±6\sqrt{7}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-14±6\sqrt{7}}{4}$$

Now solve the equation x=\frac{-14±6\sqrt{7}}{4} when ± is plus. Add -14 to 6\sqrt{7}.

$$x=\frac{6\sqrt{7}-14}{4}$$

Divide -14+6\sqrt{7} by 4.

$$x=\frac{3\sqrt{7}-7}{2}$$

Now solve the equation x=\frac{-14±6\sqrt{7}}{4} when ± is minus. Subtract 6\sqrt{7} from -14.

$$x=\frac{-6\sqrt{7}-14}{4}$$

Divide -14-6\sqrt{7} by 4.

$$x=\frac{-3\sqrt{7}-7}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{7}-7}{2} x=\frac{-3\sqrt{7}-7}{2}$$

Steps for Completing the Square

Calculate 6 to the power of 2 and get 36.

$$36=\left(5+x\right)^{2}+\left(x+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+x\right)^{2}.

$$36=25+10x+x^{2}+\left(x+2\right)^{2}$$

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

$$36=25+10x+x^{2}+x^{2}+4x+4$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$36=25+10x+2x^{2}+4x+4$$

Combine 10x and 4x to get 14x.

$$36=25+14x+2x^{2}+4$$

Add 25 and 4 to get 29.

$$36=29+14x+2x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$29+14x+2x^{2}=36$$

Subtract 29 from both sides.

$$14x+2x^{2}=36-29$$

Subtract 29 from 36 to get 7.

$$14x+2x^{2}=7$$

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.

$$2x^{2}+14x=7$$

Divide both sides by 2.

$$\frac{2x^{2}+14x}{2}=\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{14}{2}x=\frac{7}{2}$$

Divide 14 by 2.

$$x^{2}+7x=\frac{7}{2}$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=\frac{7}{2}+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=\frac{7}{2}+\frac{49}{4}$$

Add \frac{7}{2} to \frac{49}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+7x+\frac{49}{4}=\frac{63}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{63}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{63}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{3\sqrt{7}}{2} x+\frac{7}{2}=-\frac{3\sqrt{7}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=\frac{3\sqrt{7}-7}{2} x=\frac{-3\sqrt{7}-7}{2}$$