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4x^{2}-115x+350=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(-115\right)±\sqrt{\left(-115\right)^{2}-4\times 4\times 350}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -115 for b, and 350 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-115\right)±\sqrt{13225-4\times 4\times 350}}{2\times 4}
Square -115.
x=\frac{-\left(-115\right)±\sqrt{13225-16\times 350}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-115\right)±\sqrt{13225-5600}}{2\times 4}
Multiply -16 times 350.
x=\frac{-\left(-115\right)±\sqrt{7625}}{2\times 4}
Add 13225 to -5600.
x=\frac{-\left(-115\right)±5\sqrt{305}}{2\times 4}
Take the square root of 7625.
x=\frac{115±5\sqrt{305}}{2\times 4}
The opposite of -115 is 115.
x=\frac{115±5\sqrt{305}}{8}
Multiply 2 times 4.
x=\frac{5\sqrt{305}+115}{8}
Now solve the equation x=\frac{115±5\sqrt{305}}{8} when ± is plus. Add 115 to 5\sqrt{305}.
x=\frac{115-5\sqrt{305}}{8}
Now solve the equation x=\frac{115±5\sqrt{305}}{8} when ± is minus. Subtract 5\sqrt{305} from 115.
x=\frac{5\sqrt{305}+115}{8} x=\frac{115-5\sqrt{305}}{8}
The equation is now solved.
4x^{2}-115x+350=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.
4x^{2}-115x+350-350=-350
Subtract 350 from both sides of the equation.
4x^{2}-115x=-350
Subtracting 350 from itself leaves 0.
\frac{4x^{2}-115x}{4}=-\frac{350}{4}
Divide both sides by 4.
x^{2}-\frac{115}{4}x=-\frac{350}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{115}{4}x=-\frac{175}{2}
Reduce the fraction \frac{-350}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{115}{4}x+\left(-\frac{115}{8}\right)^{2}=-\frac{175}{2}+\left(-\frac{115}{8}\right)^{2}
Divide -\frac{115}{4}, the coefficient of the x term, by 2 to get -\frac{115}{8}. Then add the square of -\frac{115}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{115}{4}x+\frac{13225}{64}=-\frac{175}{2}+\frac{13225}{64}
Square -\frac{115}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{115}{4}x+\frac{13225}{64}=\frac{7625}{64}
Add -\frac{175}{2} to \frac{13225}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{115}{8}\right)^{2}=\frac{7625}{64}
Factor x^{2}-\frac{115}{4}x+\frac{13225}{64}. 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{115}{8}\right)^{2}}=\sqrt{\frac{7625}{64}}
Take the square root of both sides of the equation.
x-\frac{115}{8}=\frac{5\sqrt{305}}{8} x-\frac{115}{8}=-\frac{5\sqrt{305}}{8}
Simplify.
x=\frac{5\sqrt{305}+115}{8} x=\frac{115-5\sqrt{305}}{8}
Add \frac{115}{8} to both sides of the equation.