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2x^{2}+6x-4=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{-6±\sqrt{6^{2}-4\times 2\left(-4\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 2\left(-4\right)}}{2\times 2}
Square 6.
x=\frac{-6±\sqrt{36-8\left(-4\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-6±\sqrt{36+32}}{2\times 2}
Multiply -8 times -4.
x=\frac{-6±\sqrt{68}}{2\times 2}
Add 36 to 32.
x=\frac{-6±2\sqrt{17}}{2\times 2}
Take the square root of 68.
x=\frac{-6±2\sqrt{17}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{17}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{17}}{4} when ± is plus. Add -6 to 2\sqrt{17}.
x=\frac{\sqrt{17}-3}{2}
Divide -6+2\sqrt{17} by 4.
x=\frac{-2\sqrt{17}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{17}}{4} when ± is minus. Subtract 2\sqrt{17} from -6.
x=\frac{-\sqrt{17}-3}{2}
Divide -6-2\sqrt{17} by 4.
x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}
The equation is now solved.
2x^{2}+6x-4=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}+6x-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
2x^{2}+6x=-\left(-4\right)
Subtracting -4 from itself leaves 0.
2x^{2}+6x=4
Subtract -4 from 0.
\frac{2x^{2}+6x}{2}=\frac{4}{2}
Divide both sides by 2.
x^{2}+\frac{6}{2}x=\frac{4}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+3x=\frac{4}{2}
Divide 6 by 2.
x^{2}+3x=2
Divide 4 by 2.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=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}=2+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=\frac{17}{4}
Add 2 to \frac{9}{4}.
\left(x+\frac{3}{2}\right)^{2}=\frac{17}{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{17}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{\sqrt{17}}{2} x+\frac{3}{2}=-\frac{\sqrt{17}}{2}
Simplify.
x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}
Subtract \frac{3}{2} from both sides of the equation.