Solve 2=6(5+a)^2-6*5^2

$$2 = 6 ( 5 + a ) ^ { 2 } - 6 \times 5 ^ { 2 }$$

Solve for a

$$a=\frac{2\sqrt{57}}{3}-5\approx 0.033222957$$

$$a=-\frac{2\sqrt{57}}{3}-5\approx -10.033222957$$

Steps Using the Quadratic Formula

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

$$2=6\left(25+10a+a^{2}\right)-6\times 5^{2}$$

Use the distributive property to multiply 6 by 25+10a+a^{2}.

$$2=150+60a+6a^{2}-6\times 5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$2=150+60a+6a^{2}-6\times 25$$

Multiply 6 and 25 to get 150.

$$2=150+60a+6a^{2}-150$$

Subtract 150 from 150 to get 0.

$$2=60a+6a^{2}$$

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

$$60a+6a^{2}=2$$

Subtract 2 from both sides.

$$60a+6a^{2}-2=0$$

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.

$$6a^{2}+60a-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 60 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-60±\sqrt{60^{2}-4\times 6\left(-2\right)}}{2\times 6}$$

Square 60.

$$a=\frac{-60±\sqrt{3600-4\times 6\left(-2\right)}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-60±\sqrt{3600-24\left(-2\right)}}{2\times 6}$$

Multiply -24 times -2.

$$a=\frac{-60±\sqrt{3600+48}}{2\times 6}$$

Add 3600 to 48.

$$a=\frac{-60±\sqrt{3648}}{2\times 6}$$

Take the square root of 3648.

$$a=\frac{-60±8\sqrt{57}}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{-60±8\sqrt{57}}{12}$$

Now solve the equation a=\frac{-60±8\sqrt{57}}{12} when ± is plus. Add -60 to 8\sqrt{57}.

$$a=\frac{8\sqrt{57}-60}{12}$$

Divide -60+8\sqrt{57} by 12.

$$a=\frac{2\sqrt{57}}{3}-5$$

Now solve the equation a=\frac{-60±8\sqrt{57}}{12} when ± is minus. Subtract 8\sqrt{57} from -60.

$$a=\frac{-8\sqrt{57}-60}{12}$$

Divide -60-8\sqrt{57} by 12.

$$a=-\frac{2\sqrt{57}}{3}-5$$

The equation is now solved.

$$a=\frac{2\sqrt{57}}{3}-5 a=-\frac{2\sqrt{57}}{3}-5$$

Steps for Completing the Square

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

$$2=6\left(25+10a+a^{2}\right)-6\times 5^{2}$$

Use the distributive property to multiply 6 by 25+10a+a^{2}.

$$2=150+60a+6a^{2}-6\times 5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$2=150+60a+6a^{2}-6\times 25$$

Multiply 6 and 25 to get 150.

$$2=150+60a+6a^{2}-150$$

Subtract 150 from 150 to get 0.

$$2=60a+6a^{2}$$

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

$$60a+6a^{2}=2$$

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.

$$6a^{2}+60a=2$$

Divide both sides by 6.

$$\frac{6a^{2}+60a}{6}=\frac{2}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$a^{2}+\frac{60}{6}a=\frac{2}{6}$$

Divide 60 by 6.

$$a^{2}+10a=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$a^{2}+10a=\frac{1}{3}$$

Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+10a+5^{2}=\frac{1}{3}+5^{2}$$

Square 5.

$$a^{2}+10a+25=\frac{1}{3}+25$$

Add \frac{1}{3} to 25.

$$a^{2}+10a+25=\frac{76}{3}$$

Factor a^{2}+10a+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+5\right)^{2}=\frac{76}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+5\right)^{2}}=\sqrt{\frac{76}{3}}$$

Simplify.

$$a+5=\frac{2\sqrt{57}}{3} a+5=-\frac{2\sqrt{57}}{3}$$

Subtract 5 from both sides of the equation.

$$a=\frac{2\sqrt{57}}{3}-5 a=-\frac{2\sqrt{57}}{3}-5$$