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5x^{2}-25x=0
Subtract 25x from both sides.
x\left(5x-25\right)=0
Factor out x.
x=0 x=5
To find equation solutions, solve x=0 and 5x-25=0.
5x^{2}-25x=0
Subtract 25x from both sides.
x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -25 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-25\right)±25}{2\times 5}
Take the square root of \left(-25\right)^{2}.
x=\frac{25±25}{2\times 5}
The opposite of -25 is 25.
x=\frac{25±25}{10}
Multiply 2 times 5.
x=\frac{50}{10}
Now solve the equation x=\frac{25±25}{10} when ± is plus. Add 25 to 25.
x=5
Divide 50 by 10.
x=\frac{0}{10}
Now solve the equation x=\frac{25±25}{10} when ± is minus. Subtract 25 from 25.
x=0
Divide 0 by 10.
x=5 x=0
The equation is now solved.
5x^{2}-25x=0
Subtract 25x from both sides.
\frac{5x^{2}-25x}{5}=\frac{0}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{25}{5}\right)x=\frac{0}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-5x=\frac{0}{5}
Divide -25 by 5.
x^{2}-5x=0
Divide 0 by 5.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-5x+\frac{25}{4}=\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}
Factor x^{2}-5x+\frac{25}{4}. 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{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}
Simplify.
x=5 x=0
Add \frac{5}{2} to both sides of the equation.