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4x^{2}-5x+10=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 10}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -5 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 4\times 10}}{2\times 4}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-16\times 10}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-5\right)±\sqrt{25-160}}{2\times 4}
Multiply -16 times 10.
x=\frac{-\left(-5\right)±\sqrt{-135}}{2\times 4}
Add 25 to -160.
x=\frac{-\left(-5\right)±3\sqrt{15}i}{2\times 4}
Take the square root of -135.
x=\frac{5±3\sqrt{15}i}{2\times 4}
The opposite of -5 is 5.
x=\frac{5±3\sqrt{15}i}{8}
Multiply 2 times 4.
x=\frac{5+3\sqrt{15}i}{8}
Now solve the equation x=\frac{5±3\sqrt{15}i}{8} when ± is plus. Add 5 to 3i\sqrt{15}.
x=\frac{-3\sqrt{15}i+5}{8}
Now solve the equation x=\frac{5±3\sqrt{15}i}{8} when ± is minus. Subtract 3i\sqrt{15} from 5.
x=\frac{5+3\sqrt{15}i}{8} x=\frac{-3\sqrt{15}i+5}{8}
The equation is now solved.
4x^{2}-5x+10=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.
4x^{2}-5x+10-10=-10
Subtract 10 from both sides of the equation.
4x^{2}-5x=-10
Subtracting 10 from itself leaves 0.
\frac{4x^{2}-5x}{4}=-\frac{10}{4}
Divide both sides by 4.
x^{2}-\frac{5}{4}x=-\frac{10}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-\frac{5}{4}x=-\frac{5}{2}
Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{5}{2}+\left(-\frac{5}{8}\right)^{2}
Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{5}{2}+\frac{25}{64}
Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{4}x+\frac{25}{64}=-\frac{135}{64}
Add -\frac{5}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{8}\right)^{2}=-\frac{135}{64}
Factor x^{2}-\frac{5}{4}x+\frac{25}{64}. 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}{8}\right)^{2}}=\sqrt{-\frac{135}{64}}
Take the square root of both sides of the equation.
x-\frac{5}{8}=\frac{3\sqrt{15}i}{8} x-\frac{5}{8}=-\frac{3\sqrt{15}i}{8}
Simplify.
x=\frac{5+3\sqrt{15}i}{8} x=\frac{-3\sqrt{15}i+5}{8}
Add \frac{5}{8} to both sides of the equation.