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5x^{2}-48x+20=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(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 5\times 20}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -48 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-48\right)±\sqrt{2304-4\times 5\times 20}}{2\times 5}
Square -48.
x=\frac{-\left(-48\right)±\sqrt{2304-20\times 20}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-48\right)±\sqrt{2304-400}}{2\times 5}
Multiply -20 times 20.
x=\frac{-\left(-48\right)±\sqrt{1904}}{2\times 5}
Add 2304 to -400.
x=\frac{-\left(-48\right)±4\sqrt{119}}{2\times 5}
Take the square root of 1904.
x=\frac{48±4\sqrt{119}}{2\times 5}
The opposite of -48 is 48.
x=\frac{48±4\sqrt{119}}{10}
Multiply 2 times 5.
x=\frac{4\sqrt{119}+48}{10}
Now solve the equation x=\frac{48±4\sqrt{119}}{10} when ± is plus. Add 48 to 4\sqrt{119}.
x=\frac{2\sqrt{119}+24}{5}
Divide 48+4\sqrt{119} by 10.
x=\frac{48-4\sqrt{119}}{10}
Now solve the equation x=\frac{48±4\sqrt{119}}{10} when ± is minus. Subtract 4\sqrt{119} from 48.
x=\frac{24-2\sqrt{119}}{5}
Divide 48-4\sqrt{119} by 10.
x=\frac{2\sqrt{119}+24}{5} x=\frac{24-2\sqrt{119}}{5}
The equation is now solved.
5x^{2}-48x+20=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}-48x+20-20=-20
Subtract 20 from both sides of the equation.
5x^{2}-48x=-20
Subtracting 20 from itself leaves 0.
\frac{5x^{2}-48x}{5}=-\frac{20}{5}
Divide both sides by 5.
x^{2}-\frac{48}{5}x=-\frac{20}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{48}{5}x=-4
Divide -20 by 5.
x^{2}-\frac{48}{5}x+\left(-\frac{24}{5}\right)^{2}=-4+\left(-\frac{24}{5}\right)^{2}
Divide -\frac{48}{5}, the coefficient of the x term, by 2 to get -\frac{24}{5}. Then add the square of -\frac{24}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{48}{5}x+\frac{576}{25}=-4+\frac{576}{25}
Square -\frac{24}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{48}{5}x+\frac{576}{25}=\frac{476}{25}
Add -4 to \frac{576}{25}.
\left(x-\frac{24}{5}\right)^{2}=\frac{476}{25}
Factor x^{2}-\frac{48}{5}x+\frac{576}{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{24}{5}\right)^{2}}=\sqrt{\frac{476}{25}}
Take the square root of both sides of the equation.
x-\frac{24}{5}=\frac{2\sqrt{119}}{5} x-\frac{24}{5}=-\frac{2\sqrt{119}}{5}
Simplify.
x=\frac{2\sqrt{119}+24}{5} x=\frac{24-2\sqrt{119}}{5}
Add \frac{24}{5} to both sides of the equation.