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-2x^{2}+10x-6=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{-10±\sqrt{10^{2}-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 10 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\left(-2\right)\left(-6\right)}}{2\left(-2\right)}
Square 10.
x=\frac{-10±\sqrt{100+8\left(-6\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-10±\sqrt{100-48}}{2\left(-2\right)}
Multiply 8 times -6.
x=\frac{-10±\sqrt{52}}{2\left(-2\right)}
Add 100 to -48.
x=\frac{-10±2\sqrt{13}}{2\left(-2\right)}
Take the square root of 52.
x=\frac{-10±2\sqrt{13}}{-4}
Multiply 2 times -2.
x=\frac{2\sqrt{13}-10}{-4}
Now solve the equation x=\frac{-10±2\sqrt{13}}{-4} when ± is plus. Add -10 to 2\sqrt{13}.
x=\frac{5-\sqrt{13}}{2}
Divide -10+2\sqrt{13} by -4.
x=\frac{-2\sqrt{13}-10}{-4}
Now solve the equation x=\frac{-10±2\sqrt{13}}{-4} when ± is minus. Subtract 2\sqrt{13} from -10.
x=\frac{\sqrt{13}+5}{2}
Divide -10-2\sqrt{13} by -4.
x=\frac{5-\sqrt{13}}{2} x=\frac{\sqrt{13}+5}{2}
The equation is now solved.
-2x^{2}+10x-6=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.
-2x^{2}+10x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
-2x^{2}+10x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
-2x^{2}+10x=6
Subtract -6 from 0.
\frac{-2x^{2}+10x}{-2}=\frac{6}{-2}
Divide both sides by -2.
x^{2}+\frac{10}{-2}x=\frac{6}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}-5x=\frac{6}{-2}
Divide 10 by -2.
x^{2}-5x=-3
Divide 6 by -2.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-3+\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}=-3+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=\frac{13}{4}
Add -3 to \frac{25}{4}.
\left(x-\frac{5}{2}\right)^{2}=\frac{13}{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{13}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{\sqrt{13}}{2} x-\frac{5}{2}=-\frac{\sqrt{13}}{2}
Simplify.
x=\frac{\sqrt{13}+5}{2} x=\frac{5-\sqrt{13}}{2}
Add \frac{5}{2} to both sides of the equation.