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3x-6x^{2}=0
Subtract 6x^{2} from both sides.
x\left(3-6x\right)=0
Factor out x.
x=0 x=\frac{1}{2}
To find equation solutions, solve x=0 and 3-6x=0.
3x-6x^{2}=0
Subtract 6x^{2} from both sides.
-6x^{2}+3x=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{-3±\sqrt{3^{2}}}{2\left(-6\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±3}{2\left(-6\right)}
Take the square root of 3^{2}.
x=\frac{-3±3}{-12}
Multiply 2 times -6.
x=\frac{0}{-12}
Now solve the equation x=\frac{-3±3}{-12} when ± is plus. Add -3 to 3.
x=0
Divide 0 by -12.
x=-\frac{6}{-12}
Now solve the equation x=\frac{-3±3}{-12} when ± is minus. Subtract 3 from -3.
x=\frac{1}{2}
Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.
x=0 x=\frac{1}{2}
The equation is now solved.
3x-6x^{2}=0
Subtract 6x^{2} from both sides.
-6x^{2}+3x=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.
\frac{-6x^{2}+3x}{-6}=\frac{0}{-6}
Divide both sides by -6.
x^{2}+\frac{3}{-6}x=\frac{0}{-6}
Dividing by -6 undoes the multiplication by -6.
x^{2}-\frac{1}{2}x=\frac{0}{-6}
Reduce the fraction \frac{3}{-6} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{1}{2}x=0
Divide 0 by -6.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}
Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}
Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. 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{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}
Simplify.
x=\frac{1}{2} x=0
Add \frac{1}{4} to both sides of the equation.