Solve 1/4a^2+3a+9=4a^2

$$\frac { 1 } { 4 } a ^ { 2 } + 3 a + 9 = 4 a ^ { 2 }$$

Solve for a

$$a = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

$$a=2$$

Steps Using the Quadratic Formula

Subtract 4a^{2} from both sides.

$$\frac{1}{4}a^{2}+3a+9-4a^{2}=0$$

Combine \frac{1}{4}a^{2} and -4a^{2} to get -\frac{15}{4}a^{2}.

$$-\frac{15}{4}a^{2}+3a+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{15}{4} for a, 3 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-3±\sqrt{3^{2}-4\left(-\frac{15}{4}\right)\times 9}}{2\left(-\frac{15}{4}\right)}$$

Square 3.

$$a=\frac{-3±\sqrt{9-4\left(-\frac{15}{4}\right)\times 9}}{2\left(-\frac{15}{4}\right)}$$

Multiply -4 times -\frac{15}{4}.

$$a=\frac{-3±\sqrt{9+15\times 9}}{2\left(-\frac{15}{4}\right)}$$

Multiply 15 times 9.

$$a=\frac{-3±\sqrt{9+135}}{2\left(-\frac{15}{4}\right)}$$

Add 9 to 135.

$$a=\frac{-3±\sqrt{144}}{2\left(-\frac{15}{4}\right)}$$

Take the square root of 144.

$$a=\frac{-3±12}{2\left(-\frac{15}{4}\right)}$$

Multiply 2 times -\frac{15}{4}.

$$a=\frac{-3±12}{-\frac{15}{2}}$$

Now solve the equation a=\frac{-3±12}{-\frac{15}{2}} when ± is plus. Add -3 to 12.

$$a=\frac{9}{-\frac{15}{2}}$$

Divide 9 by -\frac{15}{2} by multiplying 9 by the reciprocal of -\frac{15}{2}.

$$a=-\frac{6}{5}$$

Now solve the equation a=\frac{-3±12}{-\frac{15}{2}} when ± is minus. Subtract 12 from -3.

$$a=-\frac{15}{-\frac{15}{2}}$$

Divide -15 by -\frac{15}{2} by multiplying -15 by the reciprocal of -\frac{15}{2}.

$$a=2$$

The equation is now solved.

$$a=-\frac{6}{5} a=2$$

Steps for Completing the Square

Subtract 4a^{2} from both sides.

$$\frac{1}{4}a^{2}+3a+9-4a^{2}=0$$

Combine \frac{1}{4}a^{2} and -4a^{2} to get -\frac{15}{4}a^{2}.

$$-\frac{15}{4}a^{2}+3a+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-\frac{15}{4}a^{2}+3a=-9$$

Divide both sides of the equation by -\frac{15}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{15}{4}a^{2}+3a}{-\frac{15}{4}}=-\frac{9}{-\frac{15}{4}}$$

Dividing by -\frac{15}{4} undoes the multiplication by -\frac{15}{4}.

$$a^{2}+\frac{3}{-\frac{15}{4}}a=-\frac{9}{-\frac{15}{4}}$$

Divide 3 by -\frac{15}{4} by multiplying 3 by the reciprocal of -\frac{15}{4}.

$$a^{2}-\frac{4}{5}a=-\frac{9}{-\frac{15}{4}}$$

Divide -9 by -\frac{15}{4} by multiplying -9 by the reciprocal of -\frac{15}{4}.

$$a^{2}-\frac{4}{5}a=\frac{12}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{4}{5}a+\left(-\frac{2}{5}\right)^{2}=\frac{12}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{4}{5}a+\frac{4}{25}=\frac{12}{5}+\frac{4}{25}$$

Add \frac{12}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{4}{5}a+\frac{4}{25}=\frac{64}{25}$$

Factor a^{2}-\frac{4}{5}a+\frac{4}{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}.

$$\left(a-\frac{2}{5}\right)^{2}=\frac{64}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{2}{5}\right)^{2}}=\sqrt{\frac{64}{25}}$$

Simplify.

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

Add \frac{2}{5} to both sides of the equation.

$$a=2 a=-\frac{6}{5}$$