Solve for x (complex solution)
x=\frac{\sqrt{15}i}{40}+\frac{5}{8}\approx 0.625+0.096824584i
x=-\frac{\sqrt{15}i}{40}+\frac{5}{8}\approx 0.625-0.096824584i
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80x^{2}-100x+32=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(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 80\times 32}}{2\times 80}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 80 for a, -100 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-100\right)±\sqrt{10000-4\times 80\times 32}}{2\times 80}
Square -100.
x=\frac{-\left(-100\right)±\sqrt{10000-320\times 32}}{2\times 80}
Multiply -4 times 80.
x=\frac{-\left(-100\right)±\sqrt{10000-10240}}{2\times 80}
Multiply -320 times 32.
x=\frac{-\left(-100\right)±\sqrt{-240}}{2\times 80}
Add 10000 to -10240.
x=\frac{-\left(-100\right)±4\sqrt{15}i}{2\times 80}
Take the square root of -240.
x=\frac{100±4\sqrt{15}i}{2\times 80}
The opposite of -100 is 100.
x=\frac{100±4\sqrt{15}i}{160}
Multiply 2 times 80.
x=\frac{100+4\sqrt{15}i}{160}
Now solve the equation x=\frac{100±4\sqrt{15}i}{160} when ± is plus. Add 100 to 4i\sqrt{15}.
x=\frac{\sqrt{15}i}{40}+\frac{5}{8}
Divide 100+4i\sqrt{15} by 160.
x=\frac{-4\sqrt{15}i+100}{160}
Now solve the equation x=\frac{100±4\sqrt{15}i}{160} when ± is minus. Subtract 4i\sqrt{15} from 100.
x=-\frac{\sqrt{15}i}{40}+\frac{5}{8}
Divide 100-4i\sqrt{15} by 160.
x=\frac{\sqrt{15}i}{40}+\frac{5}{8} x=-\frac{\sqrt{15}i}{40}+\frac{5}{8}
The equation is now solved.
80x^{2}-100x+32=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.
80x^{2}-100x+32-32=-32
Subtract 32 from both sides of the equation.
80x^{2}-100x=-32
Subtracting 32 from itself leaves 0.
\frac{80x^{2}-100x}{80}=-\frac{32}{80}
Divide both sides by 80.
x^{2}+\left(-\frac{100}{80}\right)x=-\frac{32}{80}
Dividing by 80 undoes the multiplication by 80.
x^{2}-\frac{5}{4}x=-\frac{32}{80}
Reduce the fraction \frac{-100}{80} to lowest terms by extracting and canceling out 20.
x^{2}-\frac{5}{4}x=-\frac{2}{5}
Reduce the fraction \frac{-32}{80} to lowest terms by extracting and canceling out 16.
x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=-\frac{2}{5}+\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{2}{5}+\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{3}{320}
Add -\frac{2}{5} 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{3}{320}
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{3}{320}}
Take the square root of both sides of the equation.
x-\frac{5}{8}=\frac{\sqrt{15}i}{40} x-\frac{5}{8}=-\frac{\sqrt{15}i}{40}
Simplify.
x=\frac{\sqrt{15}i}{40}+\frac{5}{8} x=-\frac{\sqrt{15}i}{40}+\frac{5}{8}
Add \frac{5}{8} to both sides of the equation.
x ^ 2 -\frac{5}{4}x +\frac{2}{5} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 80
r + s = \frac{5}{4} rs = \frac{2}{5}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{5}{8} - u s = \frac{5}{8} + u
Two numbers r and s sum up to \frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{4} = \frac{5}{8}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{5}{8} - u) (\frac{5}{8} + u) = \frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}
\frac{25}{64} - u^2 = \frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{5}-\frac{25}{64} = \frac{3}{320}
Simplify the expression by subtracting \frac{25}{64} on both sides
u^2 = -\frac{3}{320} u = \pm\sqrt{-\frac{3}{320}} = \pm \frac{\sqrt{3}}{\sqrt{320}}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{8} - \frac{\sqrt{3}}{\sqrt{320}}i = 0.625 - 0.097i s = \frac{5}{8} + \frac{\sqrt{3}}{\sqrt{320}}i = 0.625 + 0.097i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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