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100x^{2}-50x+18=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(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 100\times 18}}{2\times 100}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, -50 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-50\right)±\sqrt{2500-4\times 100\times 18}}{2\times 100}
Square -50.
x=\frac{-\left(-50\right)±\sqrt{2500-400\times 18}}{2\times 100}
Multiply -4 times 100.
x=\frac{-\left(-50\right)±\sqrt{2500-7200}}{2\times 100}
Multiply -400 times 18.
x=\frac{-\left(-50\right)±\sqrt{-4700}}{2\times 100}
Add 2500 to -7200.
x=\frac{-\left(-50\right)±10\sqrt{47}i}{2\times 100}
Take the square root of -4700.
x=\frac{50±10\sqrt{47}i}{2\times 100}
The opposite of -50 is 50.
x=\frac{50±10\sqrt{47}i}{200}
Multiply 2 times 100.
x=\frac{50+10\sqrt{47}i}{200}
Now solve the equation x=\frac{50±10\sqrt{47}i}{200} when ± is plus. Add 50 to 10i\sqrt{47}.
x=\frac{\sqrt{47}i}{20}+\frac{1}{4}
Divide 50+10i\sqrt{47} by 200.
x=\frac{-10\sqrt{47}i+50}{200}
Now solve the equation x=\frac{50±10\sqrt{47}i}{200} when ± is minus. Subtract 10i\sqrt{47} from 50.
x=-\frac{\sqrt{47}i}{20}+\frac{1}{4}
Divide 50-10i\sqrt{47} by 200.
x=\frac{\sqrt{47}i}{20}+\frac{1}{4} x=-\frac{\sqrt{47}i}{20}+\frac{1}{4}
The equation is now solved.
100x^{2}-50x+18=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.
100x^{2}-50x+18-18=-18
Subtract 18 from both sides of the equation.
100x^{2}-50x=-18
Subtracting 18 from itself leaves 0.
\frac{100x^{2}-50x}{100}=-\frac{18}{100}
Divide both sides by 100.
x^{2}+\left(-\frac{50}{100}\right)x=-\frac{18}{100}
Dividing by 100 undoes the multiplication by 100.
x^{2}-\frac{1}{2}x=-\frac{18}{100}
Reduce the fraction \frac{-50}{100} to lowest terms by extracting and canceling out 50.
x^{2}-\frac{1}{2}x=-\frac{9}{50}
Reduce the fraction \frac{-18}{100} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{9}{50}+\left(-\frac{1}{4}\right)^{2}
Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{9}{50}+\frac{1}{16}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{47}{400}
Add -\frac{9}{50} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{4}\right)^{2}=-\frac{47}{400}
Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. 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{1}{4}\right)^{2}}=\sqrt{-\frac{47}{400}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{\sqrt{47}i}{20} x-\frac{1}{4}=-\frac{\sqrt{47}i}{20}
Simplify.
x=\frac{\sqrt{47}i}{20}+\frac{1}{4} x=-\frac{\sqrt{47}i}{20}+\frac{1}{4}
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x +\frac{9}{50} = 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 100
r + s = \frac{1}{2} rs = \frac{9}{50}
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{1}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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{1}{4} - u) (\frac{1}{4} + u) = \frac{9}{50}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{50}
\frac{1}{16} - u^2 = \frac{9}{50}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{9}{50}-\frac{1}{16} = \frac{47}{400}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = -\frac{47}{400} u = \pm\sqrt{-\frac{47}{400}} = \pm \frac{\sqrt{47}}{20}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{1}{4} - \frac{\sqrt{47}}{20}i = 0.250 - 0.343i s = \frac{1}{4} + \frac{\sqrt{47}}{20}i = 0.250 + 0.343i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.