Solve a^2+12a/5=3

$$a ^ { 2 } + \frac { 12 a } { 5 } = 3$$

Solve for a

$$a=\frac{\sqrt{111}-6}{5}\approx 0.907130751$$

$$a=\frac{-\sqrt{111}-6}{5}\approx -3.307130751$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$5a^{2}+12a=15$$

Subtract 15 from both sides.

$$5a^{2}+12a-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 12 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-12±\sqrt{12^{2}-4\times 5\left(-15\right)}}{2\times 5}$$

Square 12.

$$a=\frac{-12±\sqrt{144-4\times 5\left(-15\right)}}{2\times 5}$$

Multiply -4 times 5.

$$a=\frac{-12±\sqrt{144-20\left(-15\right)}}{2\times 5}$$

Multiply -20 times -15.

$$a=\frac{-12±\sqrt{144+300}}{2\times 5}$$

Add 144 to 300.

$$a=\frac{-12±\sqrt{444}}{2\times 5}$$

Take the square root of 444.

$$a=\frac{-12±2\sqrt{111}}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{-12±2\sqrt{111}}{10}$$

Now solve the equation a=\frac{-12±2\sqrt{111}}{10} when ± is plus. Add -12 to 2\sqrt{111}.

$$a=\frac{2\sqrt{111}-12}{10}$$

Divide -12+2\sqrt{111} by 10.

$$a=\frac{\sqrt{111}-6}{5}$$

Now solve the equation a=\frac{-12±2\sqrt{111}}{10} when ± is minus. Subtract 2\sqrt{111} from -12.

$$a=\frac{-2\sqrt{111}-12}{10}$$

Divide -12-2\sqrt{111} by 10.

$$a=\frac{-\sqrt{111}-6}{5}$$

The equation is now solved.

$$a=\frac{\sqrt{111}-6}{5} a=\frac{-\sqrt{111}-6}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$5a^{2}+12a=15$$

Divide both sides by 5.

$$\frac{5a^{2}+12a}{5}=\frac{15}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$a^{2}+\frac{12}{5}a=\frac{15}{5}$$

Divide 15 by 5.

$$a^{2}+\frac{12}{5}a=3$$

Divide \frac{12}{5}, the coefficient of the x term, by 2 to get \frac{6}{5}. Then add the square of \frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{12}{5}a+\left(\frac{6}{5}\right)^{2}=3+\left(\frac{6}{5}\right)^{2}$$

Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{12}{5}a+\frac{36}{25}=3+\frac{36}{25}$$

Add 3 to \frac{36}{25}.

$$a^{2}+\frac{12}{5}a+\frac{36}{25}=\frac{111}{25}$$

Factor a^{2}+\frac{12}{5}a+\frac{36}{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}.

$$\left(a+\frac{6}{5}\right)^{2}=\frac{111}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{6}{5}\right)^{2}}=\sqrt{\frac{111}{25}}$$

Simplify.

$$a+\frac{6}{5}=\frac{\sqrt{111}}{5} a+\frac{6}{5}=-\frac{\sqrt{111}}{5}$$

Subtract \frac{6}{5} from both sides of the equation.

$$a=\frac{\sqrt{111}-6}{5} a=\frac{-\sqrt{111}-6}{5}$$