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27x^{2}-36x-20=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(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 27\left(-20\right)}}{2\times 27}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 27 for a, -36 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-36\right)±\sqrt{1296-4\times 27\left(-20\right)}}{2\times 27}
Square -36.
x=\frac{-\left(-36\right)±\sqrt{1296-108\left(-20\right)}}{2\times 27}
Multiply -4 times 27.
x=\frac{-\left(-36\right)±\sqrt{1296+2160}}{2\times 27}
Multiply -108 times -20.
x=\frac{-\left(-36\right)±\sqrt{3456}}{2\times 27}
Add 1296 to 2160.
x=\frac{-\left(-36\right)±24\sqrt{6}}{2\times 27}
Take the square root of 3456.
x=\frac{36±24\sqrt{6}}{2\times 27}
The opposite of -36 is 36.
x=\frac{36±24\sqrt{6}}{54}
Multiply 2 times 27.
x=\frac{24\sqrt{6}+36}{54}
Now solve the equation x=\frac{36±24\sqrt{6}}{54} when ± is plus. Add 36 to 24\sqrt{6}.
x=\frac{4\sqrt{6}}{9}+\frac{2}{3}
Divide 36+24\sqrt{6} by 54.
x=\frac{36-24\sqrt{6}}{54}
Now solve the equation x=\frac{36±24\sqrt{6}}{54} when ± is minus. Subtract 24\sqrt{6} from 36.
x=-\frac{4\sqrt{6}}{9}+\frac{2}{3}
Divide 36-24\sqrt{6} by 54.
x=\frac{4\sqrt{6}}{9}+\frac{2}{3} x=-\frac{4\sqrt{6}}{9}+\frac{2}{3}
The equation is now solved.
27x^{2}-36x-20=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.
27x^{2}-36x-20-\left(-20\right)=-\left(-20\right)
Add 20 to both sides of the equation.
27x^{2}-36x=-\left(-20\right)
Subtracting -20 from itself leaves 0.
27x^{2}-36x=20
Subtract -20 from 0.
\frac{27x^{2}-36x}{27}=\frac{20}{27}
Divide both sides by 27.
x^{2}+\left(-\frac{36}{27}\right)x=\frac{20}{27}
Dividing by 27 undoes the multiplication by 27.
x^{2}-\frac{4}{3}x=\frac{20}{27}
Reduce the fraction \frac{-36}{27} to lowest terms by extracting and canceling out 9.
x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{20}{27}+\left(-\frac{2}{3}\right)^{2}
Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{20}{27}+\frac{4}{9}
Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{32}{27}
Add \frac{20}{27} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{2}{3}\right)^{2}=\frac{32}{27}
Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. 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{2}{3}\right)^{2}}=\sqrt{\frac{32}{27}}
Take the square root of both sides of the equation.
x-\frac{2}{3}=\frac{4\sqrt{6}}{9} x-\frac{2}{3}=-\frac{4\sqrt{6}}{9}
Simplify.
x=\frac{4\sqrt{6}}{9}+\frac{2}{3} x=-\frac{4\sqrt{6}}{9}+\frac{2}{3}
Add \frac{2}{3} to both sides of the equation.
x ^ 2 -\frac{4}{3}x -\frac{20}{27} = 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 27
r + s = \frac{4}{3} rs = -\frac{20}{27}
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{2}{3} - u s = \frac{2}{3} + u
Two numbers r and s sum up to \frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{4}{3} = \frac{2}{3}. 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{2}{3} - u) (\frac{2}{3} + u) = -\frac{20}{27}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{27}
\frac{4}{9} - u^2 = -\frac{20}{27}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{20}{27}-\frac{4}{9} = -\frac{32}{27}
Simplify the expression by subtracting \frac{4}{9} on both sides
u^2 = \frac{32}{27} u = \pm\sqrt{\frac{32}{27}} = \pm \frac{\sqrt{32}}{\sqrt{27}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{2}{3} - \frac{\sqrt{32}}{\sqrt{27}} = -0.422 s = \frac{2}{3} + \frac{\sqrt{32}}{\sqrt{27}} = 1.755
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.