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x^{2}-3x=0
Subtract 3x from both sides.
x\left(x-3\right)=0
Factor out x.
x=0 x=3
To find equation solutions, solve x=0 and x-3=0.
x^{2}-3x=0
Subtract 3x from both sides.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±3}{2}
Take the square root of \left(-3\right)^{2}.
x=\frac{3±3}{2}
The opposite of -3 is 3.
x=\frac{6}{2}
Now solve the equation x=\frac{3±3}{2} when ± is plus. Add 3 to 3.
x=3
Divide 6 by 2.
x=\frac{0}{2}
Now solve the equation x=\frac{3±3}{2} when ± is minus. Subtract 3 from 3.
x=0
Divide 0 by 2.
x=3 x=0
The equation is now solved.
x^{2}-3x=0
Subtract 3x from both sides.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-3x+\frac{9}{4}=\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}
Factor x^{2}-3x+\frac{9}{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{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}
Simplify.
x=3 x=0
Add \frac{3}{2} to both sides of the equation.