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5x^{2}-4x+5=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{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 5\times 5}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -4 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 5\times 5}}{2\times 5}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-20\times 5}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-4\right)±\sqrt{16-100}}{2\times 5}
Multiply -20 times 5.
x=\frac{-\left(-4\right)±\sqrt{-84}}{2\times 5}
Add 16 to -100.
x=\frac{-\left(-4\right)±2\sqrt{21}i}{2\times 5}
Take the square root of -84.
x=\frac{4±2\sqrt{21}i}{2\times 5}
The opposite of -4 is 4.
x=\frac{4±2\sqrt{21}i}{10}
Multiply 2 times 5.
x=\frac{4+2\sqrt{21}i}{10}
Now solve the equation x=\frac{4±2\sqrt{21}i}{10} when ± is plus. Add 4 to 2i\sqrt{21}.
x=\frac{2+\sqrt{21}i}{5}
Divide 4+2i\sqrt{21} by 10.
x=\frac{-2\sqrt{21}i+4}{10}
Now solve the equation x=\frac{4±2\sqrt{21}i}{10} when ± is minus. Subtract 2i\sqrt{21} from 4.
x=\frac{-\sqrt{21}i+2}{5}
Divide 4-2i\sqrt{21} by 10.
x=\frac{2+\sqrt{21}i}{5} x=\frac{-\sqrt{21}i+2}{5}
The equation is now solved.
5x^{2}-4x+5=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.
5x^{2}-4x+5-5=-5
Subtract 5 from both sides of the equation.
5x^{2}-4x=-5
Subtracting 5 from itself leaves 0.
\frac{5x^{2}-4x}{5}=-\frac{5}{5}
Divide both sides by 5.
x^{2}-\frac{4}{5}x=-\frac{5}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{4}{5}x=-1
Divide -5 by 5.
x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-1+\left(-\frac{2}{5}\right)^{2}
Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-1+\frac{4}{25}
Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{5}x+\frac{4}{25}=-\frac{21}{25}
Add -1 to \frac{4}{25}.
\left(x-\frac{2}{5}\right)^{2}=-\frac{21}{25}
Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. 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{2}{5}\right)^{2}}=\sqrt{-\frac{21}{25}}
Take the square root of both sides of the equation.
x-\frac{2}{5}=\frac{\sqrt{21}i}{5} x-\frac{2}{5}=-\frac{\sqrt{21}i}{5}
Simplify.
x=\frac{2+\sqrt{21}i}{5} x=\frac{-\sqrt{21}i+2}{5}
Add \frac{2}{5} to both sides of the equation.