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\frac{1}{2}x^{2}+4x-2=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{-4±\sqrt{4^{2}-4\times \frac{1}{2}\left(-2\right)}}{2\times \frac{1}{2}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, 4 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-4±\sqrt{16-4\times \frac{1}{2}\left(-2\right)}}{2\times \frac{1}{2}}
Square 4.
x=\frac{-4±\sqrt{16-2\left(-2\right)}}{2\times \frac{1}{2}}
Multiply -4 times \frac{1}{2}.
x=\frac{-4±\sqrt{16+4}}{2\times \frac{1}{2}}
Multiply -2 times -2.
x=\frac{-4±\sqrt{20}}{2\times \frac{1}{2}}
Add 16 to 4.
x=\frac{-4±2\sqrt{5}}{2\times \frac{1}{2}}
Take the square root of 20.
x=\frac{-4±2\sqrt{5}}{1}
Multiply 2 times \frac{1}{2}.
x=\frac{2\sqrt{5}-4}{1}
Now solve the equation x=\frac{-4±2\sqrt{5}}{1} when ± is plus. Add -4 to 2\sqrt{5}.
x=2\sqrt{5}-4
Divide -4+2\sqrt{5} by 1.
x=\frac{-2\sqrt{5}-4}{1}
Now solve the equation x=\frac{-4±2\sqrt{5}}{1} when ± is minus. Subtract 2\sqrt{5} from -4.
x=-2\sqrt{5}-4
Divide -4-2\sqrt{5} by 1.
x=2\sqrt{5}-4 x=-2\sqrt{5}-4
The equation is now solved.
\frac{1}{2}x^{2}+4x-2=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{1}{2}x^{2}+4x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
\frac{1}{2}x^{2}+4x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
\frac{1}{2}x^{2}+4x=2
Subtract -2 from 0.
\frac{\frac{1}{2}x^{2}+4x}{\frac{1}{2}}=\frac{2}{\frac{1}{2}}
Multiply both sides by 2.
x^{2}+\frac{4}{\frac{1}{2}}x=\frac{2}{\frac{1}{2}}
Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.
x^{2}+8x=\frac{2}{\frac{1}{2}}
Divide 4 by \frac{1}{2} by multiplying 4 by the reciprocal of \frac{1}{2}.
x^{2}+8x=4
Divide 2 by \frac{1}{2} by multiplying 2 by the reciprocal of \frac{1}{2}.
x^{2}+8x+4^{2}=4+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=4+16
Square 4.
x^{2}+8x+16=20
Add 4 to 16.
\left(x+4\right)^{2}=20
Factor x^{2}+8x+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+4\right)^{2}}=\sqrt{20}
Take the square root of both sides of the equation.
x+4=2\sqrt{5} x+4=-2\sqrt{5}
Simplify.
x=2\sqrt{5}-4 x=-2\sqrt{5}-4
Subtract 4 from both sides of the equation.