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5x^{2}-12x+8=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(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 8}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\times 8}}{2\times 5}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-20\times 8}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-12\right)±\sqrt{144-160}}{2\times 5}
Multiply -20 times 8.
x=\frac{-\left(-12\right)±\sqrt{-16}}{2\times 5}
Add 144 to -160.
x=\frac{-\left(-12\right)±4i}{2\times 5}
Take the square root of -16.
x=\frac{12±4i}{2\times 5}
The opposite of -12 is 12.
x=\frac{12±4i}{10}
Multiply 2 times 5.
x=\frac{12+4i}{10}
Now solve the equation x=\frac{12±4i}{10} when ± is plus. Add 12 to 4i.
x=\frac{6}{5}+\frac{2}{5}i
Divide 12+4i by 10.
x=\frac{12-4i}{10}
Now solve the equation x=\frac{12±4i}{10} when ± is minus. Subtract 4i from 12.
x=\frac{6}{5}-\frac{2}{5}i
Divide 12-4i by 10.
x=\frac{6}{5}+\frac{2}{5}i x=\frac{6}{5}-\frac{2}{5}i
The equation is now solved.
5x^{2}-12x+8=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}-12x+8-8=-8
Subtract 8 from both sides of the equation.
5x^{2}-12x=-8
Subtracting 8 from itself leaves 0.
\frac{5x^{2}-12x}{5}=-\frac{8}{5}
Divide both sides by 5.
x^{2}-\frac{12}{5}x=-\frac{8}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=-\frac{8}{5}+\left(-\frac{6}{5}\right)^{2}
Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{12}{5}x+\frac{36}{25}=-\frac{8}{5}+\frac{36}{25}
Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{12}{5}x+\frac{36}{25}=-\frac{4}{25}
Add -\frac{8}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{6}{5}\right)^{2}=-\frac{4}{25}
Factor x^{2}-\frac{12}{5}x+\frac{36}{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{6}{5}\right)^{2}}=\sqrt{-\frac{4}{25}}
Take the square root of both sides of the equation.
x-\frac{6}{5}=\frac{2}{5}i x-\frac{6}{5}=-\frac{2}{5}i
Simplify.
x=\frac{6}{5}+\frac{2}{5}i x=\frac{6}{5}-\frac{2}{5}i
Add \frac{6}{5} to both sides of the equation.