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\frac{2}{5}a^{2}-\frac{4}{5}a-14=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
a=\frac{-\left(-\frac{4}{5}\right)±\sqrt{\left(-\frac{4}{5}\right)^{2}-4\times \frac{2}{5}\left(-14\right)}}{2\times \frac{2}{5}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{5} for a, -\frac{4}{5} for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-\frac{4}{5}\right)±\sqrt{\frac{16}{25}-4\times \frac{2}{5}\left(-14\right)}}{2\times \frac{2}{5}}
Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.
a=\frac{-\left(-\frac{4}{5}\right)±\sqrt{\frac{16}{25}-\frac{8}{5}\left(-14\right)}}{2\times \frac{2}{5}}
Multiply -4 times \frac{2}{5}.
a=\frac{-\left(-\frac{4}{5}\right)±\sqrt{\frac{16}{25}+\frac{112}{5}}}{2\times \frac{2}{5}}
Multiply -\frac{8}{5} times -14.
a=\frac{-\left(-\frac{4}{5}\right)±\sqrt{\frac{576}{25}}}{2\times \frac{2}{5}}
Add \frac{16}{25} to \frac{112}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
a=\frac{-\left(-\frac{4}{5}\right)±\frac{24}{5}}{2\times \frac{2}{5}}
Take the square root of \frac{576}{25}.
a=\frac{\frac{4}{5}±\frac{24}{5}}{2\times \frac{2}{5}}
The opposite of -\frac{4}{5} is \frac{4}{5}.
a=\frac{\frac{4}{5}±\frac{24}{5}}{\frac{4}{5}}
Multiply 2 times \frac{2}{5}.
a=\frac{\frac{28}{5}}{\frac{4}{5}}
Now solve the equation a=\frac{\frac{4}{5}±\frac{24}{5}}{\frac{4}{5}} when ± is plus. Add \frac{4}{5} to \frac{24}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
a=7
Divide \frac{28}{5} by \frac{4}{5} by multiplying \frac{28}{5} by the reciprocal of \frac{4}{5}.
a=-\frac{4}{\frac{4}{5}}
Now solve the equation a=\frac{\frac{4}{5}±\frac{24}{5}}{\frac{4}{5}} when ± is minus. Subtract \frac{24}{5} from \frac{4}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
a=-5
Divide -4 by \frac{4}{5} by multiplying -4 by the reciprocal of \frac{4}{5}.
a=7 a=-5
The equation is now solved.
\frac{2}{5}a^{2}-\frac{4}{5}a-14=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\frac{2}{5}a^{2}-\frac{4}{5}a-14-\left(-14\right)=-\left(-14\right)
Add 14 to both sides of the equation.
\frac{2}{5}a^{2}-\frac{4}{5}a=-\left(-14\right)
Subtracting -14 from itself leaves 0.
\frac{2}{5}a^{2}-\frac{4}{5}a=14
Subtract -14 from 0.
\frac{\frac{2}{5}a^{2}-\frac{4}{5}a}{\frac{2}{5}}=\frac{14}{\frac{2}{5}}
Divide both sides of the equation by \frac{2}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
a^{2}+\left(-\frac{\frac{4}{5}}{\frac{2}{5}}\right)a=\frac{14}{\frac{2}{5}}
Dividing by \frac{2}{5} undoes the multiplication by \frac{2}{5}.
a^{2}-2a=\frac{14}{\frac{2}{5}}
Divide -\frac{4}{5} by \frac{2}{5} by multiplying -\frac{4}{5} by the reciprocal of \frac{2}{5}.
a^{2}-2a=35
Divide 14 by \frac{2}{5} by multiplying 14 by the reciprocal of \frac{2}{5}.
a^{2}-2a+1=35+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-2a+1=36
Add 35 to 1.
\left(a-1\right)^{2}=36
Factor a^{2}-2a+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(a-1\right)^{2}}=\sqrt{36}
Take the square root of both sides of the equation.
a-1=6 a-1=-6
Simplify.
a=7 a=-5
Add 1 to both sides of the equation.