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5x^{2}-10x+\frac{117}{5}=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{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 5\times \frac{117}{5}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -10 for b, and \frac{117}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 5\times \frac{117}{5}}}{2\times 5}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-20\times \frac{117}{5}}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-10\right)±\sqrt{100-468}}{2\times 5}
Multiply -20 times \frac{117}{5}.
x=\frac{-\left(-10\right)±\sqrt{-368}}{2\times 5}
Add 100 to -468.
x=\frac{-\left(-10\right)±4\sqrt{23}i}{2\times 5}
Take the square root of -368.
x=\frac{10±4\sqrt{23}i}{2\times 5}
The opposite of -10 is 10.
x=\frac{10±4\sqrt{23}i}{10}
Multiply 2 times 5.
x=\frac{10+4\sqrt{23}i}{10}
Now solve the equation x=\frac{10±4\sqrt{23}i}{10} when ± is plus. Add 10 to 4i\sqrt{23}.
x=\frac{2\sqrt{23}i}{5}+1
Divide 10+4i\sqrt{23} by 10.
x=\frac{-4\sqrt{23}i+10}{10}
Now solve the equation x=\frac{10±4\sqrt{23}i}{10} when ± is minus. Subtract 4i\sqrt{23} from 10.
x=-\frac{2\sqrt{23}i}{5}+1
Divide 10-4i\sqrt{23} by 10.
x=\frac{2\sqrt{23}i}{5}+1 x=-\frac{2\sqrt{23}i}{5}+1
The equation is now solved.
5x^{2}-10x+\frac{117}{5}=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.
5x^{2}-10x+\frac{117}{5}-\frac{117}{5}=-\frac{117}{5}
Subtract \frac{117}{5} from both sides of the equation.
5x^{2}-10x=-\frac{117}{5}
Subtracting \frac{117}{5} from itself leaves 0.
\frac{5x^{2}-10x}{5}=-\frac{\frac{117}{5}}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{10}{5}\right)x=-\frac{\frac{117}{5}}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-2x=-\frac{\frac{117}{5}}{5}
Divide -10 by 5.
x^{2}-2x=-\frac{117}{25}
Divide -\frac{117}{5} by 5.
x^{2}-2x+1=-\frac{117}{25}+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.
x^{2}-2x+1=-\frac{92}{25}
Add -\frac{117}{25} to 1.
\left(x-1\right)^{2}=-\frac{92}{25}
Factor x^{2}-2x+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(x-1\right)^{2}}=\sqrt{-\frac{92}{25}}
Take the square root of both sides of the equation.
x-1=\frac{2\sqrt{23}i}{5} x-1=-\frac{2\sqrt{23}i}{5}
Simplify.
x=\frac{2\sqrt{23}i}{5}+1 x=-\frac{2\sqrt{23}i}{5}+1
Add 1 to both sides of the equation.