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25x^{2}-90x+82=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(-90\right)±\sqrt{\left(-90\right)^{2}-4\times 25\times 82}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -90 for b, and 82 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-90\right)±\sqrt{8100-4\times 25\times 82}}{2\times 25}
Square -90.
x=\frac{-\left(-90\right)±\sqrt{8100-100\times 82}}{2\times 25}
Multiply -4 times 25.
x=\frac{-\left(-90\right)±\sqrt{8100-8200}}{2\times 25}
Multiply -100 times 82.
x=\frac{-\left(-90\right)±\sqrt{-100}}{2\times 25}
Add 8100 to -8200.
x=\frac{-\left(-90\right)±10i}{2\times 25}
Take the square root of -100.
x=\frac{90±10i}{2\times 25}
The opposite of -90 is 90.
x=\frac{90±10i}{50}
Multiply 2 times 25.
x=\frac{90+10i}{50}
Now solve the equation x=\frac{90±10i}{50} when ± is plus. Add 90 to 10i.
x=\frac{9}{5}+\frac{1}{5}i
Divide 90+10i by 50.
x=\frac{90-10i}{50}
Now solve the equation x=\frac{90±10i}{50} when ± is minus. Subtract 10i from 90.
x=\frac{9}{5}-\frac{1}{5}i
Divide 90-10i by 50.
x=\frac{9}{5}+\frac{1}{5}i x=\frac{9}{5}-\frac{1}{5}i
The equation is now solved.
25x^{2}-90x+82=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.
25x^{2}-90x+82-82=-82
Subtract 82 from both sides of the equation.
25x^{2}-90x=-82
Subtracting 82 from itself leaves 0.
\frac{25x^{2}-90x}{25}=-\frac{82}{25}
Divide both sides by 25.
x^{2}+\left(-\frac{90}{25}\right)x=-\frac{82}{25}
Dividing by 25 undoes the multiplication by 25.
x^{2}-\frac{18}{5}x=-\frac{82}{25}
Reduce the fraction \frac{-90}{25} to lowest terms by extracting and canceling out 5.
x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{82}{25}+\left(-\frac{9}{5}\right)^{2}
Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{-82+81}{25}
Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{18}{5}x+\frac{81}{25}=-\frac{1}{25}
Add -\frac{82}{25} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{9}{5}\right)^{2}=-\frac{1}{25}
Factor x^{2}-\frac{18}{5}x+\frac{81}{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{9}{5}\right)^{2}}=\sqrt{-\frac{1}{25}}
Take the square root of both sides of the equation.
x-\frac{9}{5}=\frac{1}{5}i x-\frac{9}{5}=-\frac{1}{5}i
Simplify.
x=\frac{9}{5}+\frac{1}{5}i x=\frac{9}{5}-\frac{1}{5}i
Add \frac{9}{5} to both sides of the equation.
x ^ 2 -\frac{18}{5}x +\frac{82}{25} = 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 25
r + s = \frac{18}{5} rs = \frac{82}{25}
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{9}{5} - u s = \frac{9}{5} + u
Two numbers r and s sum up to \frac{18}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{18}{5} = \frac{9}{5}. 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{9}{5} - u) (\frac{9}{5} + u) = \frac{82}{25}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{82}{25}
\frac{81}{25} - u^2 = \frac{82}{25}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{82}{25}-\frac{81}{25} = \frac{1}{25}
Simplify the expression by subtracting \frac{81}{25} on both sides
u^2 = -\frac{1}{25} u = \pm\sqrt{-\frac{1}{25}} = \pm \frac{1}{5}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{9}{5} - \frac{1}{5}i = 1.800 - 0.200i s = \frac{9}{5} + \frac{1}{5}i = 1.800 + 0.200i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.