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-10x^{2}+12x-9=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{-12±\sqrt{12^{2}-4\left(-10\right)\left(-9\right)}}{2\left(-10\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-12±\sqrt{144-4\left(-10\right)\left(-9\right)}}{2\left(-10\right)}
Square 12.
x=\frac{-12±\sqrt{144+40\left(-9\right)}}{2\left(-10\right)}
Multiply -4 times -10.
x=\frac{-12±\sqrt{144-360}}{2\left(-10\right)}
Multiply 40 times -9.
x=\frac{-12±\sqrt{-216}}{2\left(-10\right)}
Add 144 to -360.
x=\frac{-12±6\sqrt{6}i}{2\left(-10\right)}
Take the square root of -216.
x=\frac{-12±6\sqrt{6}i}{-20}
Multiply 2 times -10.
x=\frac{-12+6\sqrt{6}i}{-20}
Now solve the equation x=\frac{-12±6\sqrt{6}i}{-20} when ± is plus. Add -12 to 6i\sqrt{6}.
x=-\frac{3\sqrt{6}i}{10}+\frac{3}{5}
Divide -12+6i\sqrt{6} by -20.
x=\frac{-6\sqrt{6}i-12}{-20}
Now solve the equation x=\frac{-12±6\sqrt{6}i}{-20} when ± is minus. Subtract 6i\sqrt{6} from -12.
x=\frac{3\sqrt{6}i}{10}+\frac{3}{5}
Divide -12-6i\sqrt{6} by -20.
x=-\frac{3\sqrt{6}i}{10}+\frac{3}{5} x=\frac{3\sqrt{6}i}{10}+\frac{3}{5}
The equation is now solved.
-10x^{2}+12x-9=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.
-10x^{2}+12x-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
-10x^{2}+12x=-\left(-9\right)
Subtracting -9 from itself leaves 0.
-10x^{2}+12x=9
Subtract -9 from 0.
\frac{-10x^{2}+12x}{-10}=\frac{9}{-10}
Divide both sides by -10.
x^{2}+\frac{12}{-10}x=\frac{9}{-10}
Dividing by -10 undoes the multiplication by -10.
x^{2}-\frac{6}{5}x=\frac{9}{-10}
Reduce the fraction \frac{12}{-10} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{6}{5}x=-\frac{9}{10}
Divide 9 by -10.
x^{2}-\frac{6}{5}x+\left(-\frac{3}{5}\right)^{2}=-\frac{9}{10}+\left(-\frac{3}{5}\right)^{2}
Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{9}{10}+\frac{9}{25}
Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{6}{5}x+\frac{9}{25}=-\frac{27}{50}
Add -\frac{9}{10} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{5}\right)^{2}=-\frac{27}{50}
Factor x^{2}-\frac{6}{5}x+\frac{9}{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{3}{5}\right)^{2}}=\sqrt{-\frac{27}{50}}
Take the square root of both sides of the equation.
x-\frac{3}{5}=\frac{3\sqrt{6}i}{10} x-\frac{3}{5}=-\frac{3\sqrt{6}i}{10}
Simplify.
x=\frac{3\sqrt{6}i}{10}+\frac{3}{5} x=-\frac{3\sqrt{6}i}{10}+\frac{3}{5}
Add \frac{3}{5} to both sides of the equation.
x ^ 2 -\frac{6}{5}x +\frac{9}{10} = 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.
r + s = \frac{6}{5} rs = \frac{9}{10}
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{3}{5} - u s = \frac{3}{5} + u
Two numbers r and s sum up to \frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{6}{5} = \frac{3}{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{3}{5} - u) (\frac{3}{5} + u) = \frac{9}{10}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{10}
\frac{9}{25} - u^2 = \frac{9}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{9}{10}-\frac{9}{25} = \frac{27}{50}
Simplify the expression by subtracting \frac{9}{25} on both sides
u^2 = -\frac{27}{50} u = \pm\sqrt{-\frac{27}{50}} = \pm \frac{\sqrt{27}}{\sqrt{50}}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{3}{5} - \frac{\sqrt{27}}{\sqrt{50}}i = 0.600 - 0.735i s = \frac{3}{5} + \frac{\sqrt{27}}{\sqrt{50}}i = 0.600 + 0.735i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.