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x^{2}+8x-\frac{5}{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{-8±\sqrt{8^{2}-4\left(-\frac{5}{6}\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and -\frac{5}{6} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-8±\sqrt{64-4\left(-\frac{5}{6}\right)}}{2}
Square 8.
x=\frac{-8±\sqrt{64+\frac{10}{3}}}{2}
Multiply -4 times -\frac{5}{6}.
x=\frac{-8±\sqrt{\frac{202}{3}}}{2}
Add 64 to \frac{10}{3}.
x=\frac{-8±\frac{\sqrt{606}}{3}}{2}
Take the square root of \frac{202}{3}.
x=\frac{\frac{\sqrt{606}}{3}-8}{2}
Now solve the equation x=\frac{-8±\frac{\sqrt{606}}{3}}{2} when ± is plus. Add -8 to \frac{\sqrt{606}}{3}.
x=\frac{\sqrt{606}}{6}-4
Divide -8+\frac{\sqrt{606}}{3} by 2.
x=\frac{-\frac{\sqrt{606}}{3}-8}{2}
Now solve the equation x=\frac{-8±\frac{\sqrt{606}}{3}}{2} when ± is minus. Subtract \frac{\sqrt{606}}{3} from -8.
x=-\frac{\sqrt{606}}{6}-4
Divide -8-\frac{\sqrt{606}}{3} by 2.
x=\frac{\sqrt{606}}{6}-4 x=-\frac{\sqrt{606}}{6}-4
The equation is now solved.
x^{2}+8x-\frac{5}{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.
x^{2}+8x-\frac{5}{6}-\left(-\frac{5}{6}\right)=-\left(-\frac{5}{6}\right)
Add \frac{5}{6} to both sides of the equation.
x^{2}+8x=-\left(-\frac{5}{6}\right)
Subtracting -\frac{5}{6} from itself leaves 0.
x^{2}+8x=\frac{5}{6}
Subtract -\frac{5}{6} from 0.
x^{2}+8x+4^{2}=\frac{5}{6}+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=\frac{5}{6}+16
Square 4.
x^{2}+8x+16=\frac{101}{6}
Add \frac{5}{6} to 16.
\left(x+4\right)^{2}=\frac{101}{6}
Factor x^{2}+8x+16. 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+4\right)^{2}}=\sqrt{\frac{101}{6}}
Take the square root of both sides of the equation.
x+4=\frac{\sqrt{606}}{6} x+4=-\frac{\sqrt{606}}{6}
Simplify.
x=\frac{\sqrt{606}}{6}-4 x=-\frac{\sqrt{606}}{6}-4
Subtract 4 from both sides of the equation.