Solve 25=(2a-5)^2+a^2

$$25 = ( 2 a - 5 ) ^ { 2 } + a ^ { 2 }$$

Solve for a

$$a=4$$

$$a=0$$

Steps Using Factoring

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

$$25=4a^{2}-20a+25+a^{2}$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$25=5a^{2}-20a+25$$

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

$$5a^{2}-20a+25=25$$

Subtract 25 from both sides.

$$5a^{2}-20a+25-25=0$$

Subtract 25 from 25 to get 0.

$$5a^{2}-20a=0$$

Factor out a.

$$a\left(5a-20\right)=0$$

To find equation solutions, solve a=0 and 5a-20=0.

$$a=0 a=4$$

Steps Using the Quadratic Formula

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

$$25=4a^{2}-20a+25+a^{2}$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$25=5a^{2}-20a+25$$

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

$$5a^{2}-20a+25=25$$

Subtract 25 from both sides.

$$5a^{2}-20a+25-25=0$$

Subtract 25 from 25 to get 0.

$$5a^{2}-20a=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -20 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}}}{2\times 5}$$

Take the square root of \left(-20\right)^{2}.

$$a=\frac{-\left(-20\right)±20}{2\times 5}$$

The opposite of -20 is 20.

$$a=\frac{20±20}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{20±20}{10}$$

Now solve the equation a=\frac{20±20}{10} when ± is plus. Add 20 to 20.

$$a=\frac{40}{10}$$

Divide 40 by 10.

$$a=4$$

Now solve the equation a=\frac{20±20}{10} when ± is minus. Subtract 20 from 20.

$$a=\frac{0}{10}$$

Divide 0 by 10.

$$a=0$$

The equation is now solved.

$$a=4 a=0$$

Steps for Completing the Square

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

$$25=4a^{2}-20a+25+a^{2}$$

Combine 4a^{2} and a^{2} to get 5a^{2}.

$$25=5a^{2}-20a+25$$

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

$$5a^{2}-20a+25=25$$

Subtract 25 from both sides.

$$5a^{2}-20a=25-25$$

Subtract 25 from 25 to get 0.

$$5a^{2}-20a=0$$

Divide both sides by 5.

$$\frac{5a^{2}-20a}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}+\left(-\frac{20}{5}\right)a=\frac{0}{5}$$

Divide -20 by 5.

$$a^{2}-4a=\frac{0}{5}$$

Divide 0 by 5.

$$a^{2}-4a=0$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-4a+\left(-2\right)^{2}=\left(-2\right)^{2}$$

Square -2.

$$a^{2}-4a+4=4$$

Factor a^{2}-4a+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}.

$$\left(a-2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$a-2=2 a-2=-2$$

Add 2 to both sides of the equation.

$$a=4 a=0$$