Solve 26=a^2-10a+25+4a^2-12a+9

$$26 = a ^ { 2 } - 10 a + 25 + 4 a ^ { 2 } - 12 a + 9$$

Solve for a

$$a=\frac{2}{5}=0.4$$

$$a=4$$

Steps Using Factoring By Grouping

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

$$26=5a^{2}-10a+25-12a+9$$

Combine -10a and -12a to get -22a.

$$26=5a^{2}-22a+25+9$$

Add 25 and 9 to get 34.

$$26=5a^{2}-22a+34$$

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

$$5a^{2}-22a+34=26$$

Subtract 26 from both sides.

$$5a^{2}-22a+34-26=0$$

Subtract 26 from 34 to get 8.

$$5a^{2}-22a+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5a^{2}+aa+ba+8. To find a and b, set up a system to be solved.

$$a+b=-22 ab=5\times 8=40$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -22.

$$a=-20 b=-2$$

Rewrite 5a^{2}-22a+8 as \left(5a^{2}-20a\right)+\left(-2a+8\right).

$$\left(5a^{2}-20a\right)+\left(-2a+8\right)$$

Factor out 5a in the first and -2 in the second group.

$$5a\left(a-4\right)-2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(5a-2\right)$$

To find equation solutions, solve a-4=0 and 5a-2=0.

$$a=4 a=\frac{2}{5}$$

Steps Using the Quadratic Formula

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

$$26=5a^{2}-10a+25-12a+9$$

Combine -10a and -12a to get -22a.

$$26=5a^{2}-22a+25+9$$

Add 25 and 9 to get 34.

$$26=5a^{2}-22a+34$$

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

$$5a^{2}-22a+34=26$$

Subtract 26 from both sides.

$$5a^{2}-22a+34-26=0$$

Subtract 26 from 34 to get 8.

$$5a^{2}-22a+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -22 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 5\times 8}}{2\times 5}$$

Square -22.

$$a=\frac{-\left(-22\right)±\sqrt{484-4\times 5\times 8}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-\left(-22\right)±\sqrt{484-20\times 8}}{2\times 5}$$

Multiply -20 times 8.

$$a=\frac{-\left(-22\right)±\sqrt{484-160}}{2\times 5}$$

Add 484 to -160.

$$a=\frac{-\left(-22\right)±\sqrt{324}}{2\times 5}$$

Take the square root of 324.

$$a=\frac{-\left(-22\right)±18}{2\times 5}$$

The opposite of -22 is 22.

$$a=\frac{22±18}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{22±18}{10}$$

Now solve the equation a=\frac{22±18}{10} when ± is plus. Add 22 to 18.

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

Divide 40 by 10.

$$a=4$$

Now solve the equation a=\frac{22±18}{10} when ± is minus. Subtract 18 from 22.

$$a=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$a=\frac{2}{5}$$

The equation is now solved.

$$a=4 a=\frac{2}{5}$$

Steps for Completing the Square

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

$$26=5a^{2}-10a+25-12a+9$$

Combine -10a and -12a to get -22a.

$$26=5a^{2}-22a+25+9$$

Add 25 and 9 to get 34.

$$26=5a^{2}-22a+34$$

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

$$5a^{2}-22a+34=26$$

Subtract 34 from both sides.

$$5a^{2}-22a=26-34$$

Subtract 34 from 26 to get -8.

$$5a^{2}-22a=-8$$

Divide both sides by 5.

$$\frac{5a^{2}-22a}{5}=-\frac{8}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}-\frac{22}{5}a=-\frac{8}{5}$$

Divide -\frac{22}{5}, the coefficient of the x term, by 2 to get -\frac{11}{5}. Then add the square of -\frac{11}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{22}{5}a+\left(-\frac{11}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{11}{5}\right)^{2}$$

Square -\frac{11}{5} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{22}{5}a+\frac{121}{25}=-\frac{8}{5}+\frac{121}{25}$$

Add -\frac{8}{5} to \frac{121}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{22}{5}a+\frac{121}{25}=\frac{81}{25}$$

Factor a^{2}-\frac{22}{5}a+\frac{121}{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}.

$$\left(a-\frac{11}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{11}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$a-\frac{11}{5}=\frac{9}{5} a-\frac{11}{5}=-\frac{9}{5}$$

Add \frac{11}{5} to both sides of the equation.

$$a=4 a=\frac{2}{5}$$