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\frac{1}{4}x^{2}+\frac{3}{5}x-6=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.
x=\frac{-\frac{3}{5}±\sqrt{\left(\frac{3}{5}\right)^{2}-4\times \frac{1}{4}\left(-6\right)}}{2\times \frac{1}{4}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, \frac{3}{5} for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}-4\times \frac{1}{4}\left(-6\right)}}{2\times \frac{1}{4}}
Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}-\left(-6\right)}}{2\times \frac{1}{4}}
Multiply -4 times \frac{1}{4}.
x=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}+6}}{2\times \frac{1}{4}}
Multiply -1 times -6.
x=\frac{-\frac{3}{5}±\sqrt{\frac{159}{25}}}{2\times \frac{1}{4}}
Add \frac{9}{25} to 6.
x=\frac{-\frac{3}{5}±\frac{\sqrt{159}}{5}}{2\times \frac{1}{4}}
Take the square root of \frac{159}{25}.
x=\frac{-\frac{3}{5}±\frac{\sqrt{159}}{5}}{\frac{1}{2}}
Multiply 2 times \frac{1}{4}.
x=\frac{\sqrt{159}-3}{\frac{1}{2}\times 5}
Now solve the equation x=\frac{-\frac{3}{5}±\frac{\sqrt{159}}{5}}{\frac{1}{2}} when ± is plus. Add -\frac{3}{5} to \frac{\sqrt{159}}{5}.
x=\frac{2\sqrt{159}-6}{5}
Divide \frac{-3+\sqrt{159}}{5} by \frac{1}{2} by multiplying \frac{-3+\sqrt{159}}{5} by the reciprocal of \frac{1}{2}.
x=\frac{-\sqrt{159}-3}{\frac{1}{2}\times 5}
Now solve the equation x=\frac{-\frac{3}{5}±\frac{\sqrt{159}}{5}}{\frac{1}{2}} when ± is minus. Subtract \frac{\sqrt{159}}{5} from -\frac{3}{5}.
x=\frac{-2\sqrt{159}-6}{5}
Divide \frac{-3-\sqrt{159}}{5} by \frac{1}{2} by multiplying \frac{-3-\sqrt{159}}{5} by the reciprocal of \frac{1}{2}.
x=\frac{2\sqrt{159}-6}{5} x=\frac{-2\sqrt{159}-6}{5}
The equation is now solved.
\frac{1}{4}x^{2}+\frac{3}{5}x-6=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{1}{4}x^{2}+\frac{3}{5}x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
\frac{1}{4}x^{2}+\frac{3}{5}x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
\frac{1}{4}x^{2}+\frac{3}{5}x=6
Subtract -6 from 0.
\frac{\frac{1}{4}x^{2}+\frac{3}{5}x}{\frac{1}{4}}=\frac{6}{\frac{1}{4}}
Multiply both sides by 4.
x^{2}+\frac{\frac{3}{5}}{\frac{1}{4}}x=\frac{6}{\frac{1}{4}}
Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.
x^{2}+\frac{12}{5}x=\frac{6}{\frac{1}{4}}
Divide \frac{3}{5} by \frac{1}{4} by multiplying \frac{3}{5} by the reciprocal of \frac{1}{4}.
x^{2}+\frac{12}{5}x=24
Divide 6 by \frac{1}{4} by multiplying 6 by the reciprocal of \frac{1}{4}.
x^{2}+\frac{12}{5}x+\left(\frac{6}{5}\right)^{2}=24+\left(\frac{6}{5}\right)^{2}
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.
x^{2}+\frac{12}{5}x+\frac{36}{25}=24+\frac{36}{25}
Square \frac{6}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{12}{5}x+\frac{36}{25}=\frac{636}{25}
Add 24 to \frac{36}{25}.
\left(x+\frac{6}{5}\right)^{2}=\frac{636}{25}
Factor x^{2}+\frac{12}{5}x+\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}.
\sqrt{\left(x+\frac{6}{5}\right)^{2}}=\sqrt{\frac{636}{25}}
Take the square root of both sides of the equation.
x+\frac{6}{5}=\frac{2\sqrt{159}}{5} x+\frac{6}{5}=-\frac{2\sqrt{159}}{5}
Simplify.
x=\frac{2\sqrt{159}-6}{5} x=\frac{-2\sqrt{159}-6}{5}
Subtract \frac{6}{5} from both sides of the equation.