Solve -21=(a-5)(a-5)

$$- 21 = ( a - 5 ) ( a - 5 )$$

Solve for a

$$a=-\sqrt{21}i+5\approx 5-4.582575695i$$

$$a=5+\sqrt{21}i\approx 5+4.582575695i$$

Steps Using the Quadratic Formula

Multiply a-5 and a-5 to get \left(a-5\right)^{2}.

$$-21=\left(a-5\right)^{2}$$

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

$$-21=a^{2}-10a+25$$

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

$$a^{2}-10a+25=-21$$

Add 21 to both sides.

$$a^{2}-10a+25+21=0$$

Add 25 and 21 to get 46.

$$a^{2}-10a+46=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 46 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 46}}{2}$$

Square -10.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\times 46}}{2}$$

Multiply -4 times 46.

$$a=\frac{-\left(-10\right)±\sqrt{100-184}}{2}$$

Add 100 to -184.

$$a=\frac{-\left(-10\right)±\sqrt{-84}}{2}$$

Take the square root of -84.

$$a=\frac{-\left(-10\right)±2\sqrt{21}i}{2}$$

The opposite of -10 is 10.

$$a=\frac{10±2\sqrt{21}i}{2}$$

Now solve the equation a=\frac{10±2\sqrt{21}i}{2} when ± is plus. Add 10 to 2i\sqrt{21}.

$$a=\frac{10+2\sqrt{21}i}{2}$$

Divide 10+2i\sqrt{21} by 2.

$$a=5+\sqrt{21}i$$

Now solve the equation a=\frac{10±2\sqrt{21}i}{2} when ± is minus. Subtract 2i\sqrt{21} from 10.

$$a=\frac{-2\sqrt{21}i+10}{2}$$

Divide 10-2i\sqrt{21} by 2.

$$a=-\sqrt{21}i+5$$

The equation is now solved.

$$a=5+\sqrt{21}i a=-\sqrt{21}i+5$$

Steps for Completing the Square

Multiply a-5 and a-5 to get \left(a-5\right)^{2}.

$$-21=\left(a-5\right)^{2}$$

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

$$-21=a^{2}-10a+25$$

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

$$a^{2}-10a+25=-21$$

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}.

$$\left(a-5\right)^{2}=-21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-5\right)^{2}}=\sqrt{-21}$$

Simplify.

$$a-5=\sqrt{21}i a-5=-\sqrt{21}i$$

Add 5 to both sides of the equation.

$$a=5+\sqrt{21}i a=-\sqrt{21}i+5$$