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206x^{2}-40x+25=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(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 206\times 25}}{2\times 206}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 206 for a, -40 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-40\right)±\sqrt{1600-4\times 206\times 25}}{2\times 206}
Square -40.
x=\frac{-\left(-40\right)±\sqrt{1600-824\times 25}}{2\times 206}
Multiply -4 times 206.
x=\frac{-\left(-40\right)±\sqrt{1600-20600}}{2\times 206}
Multiply -824 times 25.
x=\frac{-\left(-40\right)±\sqrt{-19000}}{2\times 206}
Add 1600 to -20600.
x=\frac{-\left(-40\right)±10\sqrt{190}i}{2\times 206}
Take the square root of -19000.
x=\frac{40±10\sqrt{190}i}{2\times 206}
The opposite of -40 is 40.
x=\frac{40±10\sqrt{190}i}{412}
Multiply 2 times 206.
x=\frac{40+10\sqrt{190}i}{412}
Now solve the equation x=\frac{40±10\sqrt{190}i}{412} when ± is plus. Add 40 to 10i\sqrt{190}.
x=\frac{5\sqrt{190}i}{206}+\frac{10}{103}
Divide 40+10i\sqrt{190} by 412.
x=\frac{-10\sqrt{190}i+40}{412}
Now solve the equation x=\frac{40±10\sqrt{190}i}{412} when ± is minus. Subtract 10i\sqrt{190} from 40.
x=-\frac{5\sqrt{190}i}{206}+\frac{10}{103}
Divide 40-10i\sqrt{190} by 412.
x=\frac{5\sqrt{190}i}{206}+\frac{10}{103} x=-\frac{5\sqrt{190}i}{206}+\frac{10}{103}
The equation is now solved.
206x^{2}-40x+25=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.
206x^{2}-40x+25-25=-25
Subtract 25 from both sides of the equation.
206x^{2}-40x=-25
Subtracting 25 from itself leaves 0.
\frac{206x^{2}-40x}{206}=-\frac{25}{206}
Divide both sides by 206.
x^{2}+\left(-\frac{40}{206}\right)x=-\frac{25}{206}
Dividing by 206 undoes the multiplication by 206.
x^{2}-\frac{20}{103}x=-\frac{25}{206}
Reduce the fraction \frac{-40}{206} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{20}{103}x+\left(-\frac{10}{103}\right)^{2}=-\frac{25}{206}+\left(-\frac{10}{103}\right)^{2}
Divide -\frac{20}{103}, the coefficient of the x term, by 2 to get -\frac{10}{103}. Then add the square of -\frac{10}{103} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{20}{103}x+\frac{100}{10609}=-\frac{25}{206}+\frac{100}{10609}
Square -\frac{10}{103} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{20}{103}x+\frac{100}{10609}=-\frac{2375}{21218}
Add -\frac{25}{206} to \frac{100}{10609} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{10}{103}\right)^{2}=-\frac{2375}{21218}
Factor x^{2}-\frac{20}{103}x+\frac{100}{10609}. 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{10}{103}\right)^{2}}=\sqrt{-\frac{2375}{21218}}
Take the square root of both sides of the equation.
x-\frac{10}{103}=\frac{5\sqrt{190}i}{206} x-\frac{10}{103}=-\frac{5\sqrt{190}i}{206}
Simplify.
x=\frac{5\sqrt{190}i}{206}+\frac{10}{103} x=-\frac{5\sqrt{190}i}{206}+\frac{10}{103}
Add \frac{10}{103} to both sides of the equation.
x ^ 2 -\frac{20}{103}x +\frac{25}{206} = 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 206
r + s = \frac{20}{103} rs = \frac{25}{206}
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{10}{103} - u s = \frac{10}{103} + u
Two numbers r and s sum up to \frac{20}{103} exactly when the average of the two numbers is \frac{1}{2}*\frac{20}{103} = \frac{10}{103}. 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{10}{103} - u) (\frac{10}{103} + u) = \frac{25}{206}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{206}
\frac{100}{10609} - u^2 = \frac{25}{206}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{25}{206}-\frac{100}{10609} = \frac{2375}{21218}
Simplify the expression by subtracting \frac{100}{10609} on both sides
u^2 = -\frac{2375}{21218} u = \pm\sqrt{-\frac{2375}{21218}} = \pm \frac{\sqrt{2375}}{\sqrt{21218}}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{10}{103} - \frac{\sqrt{2375}}{\sqrt{21218}}i = 0.097 - 0.335i s = \frac{10}{103} + \frac{\sqrt{2375}}{\sqrt{21218}}i = 0.097 + 0.335i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.