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2x^{2}-9x-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{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-9\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-9\right)}}{2\times 2}
Square -9.
x=\frac{-\left(-9\right)±\sqrt{81-8\left(-9\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-9\right)±\sqrt{81+72}}{2\times 2}
Multiply -8 times -9.
x=\frac{-\left(-9\right)±\sqrt{153}}{2\times 2}
Add 81 to 72.
x=\frac{-\left(-9\right)±3\sqrt{17}}{2\times 2}
Take the square root of 153.
x=\frac{9±3\sqrt{17}}{2\times 2}
The opposite of -9 is 9.
x=\frac{9±3\sqrt{17}}{4}
Multiply 2 times 2.
x=\frac{3\sqrt{17}+9}{4}
Now solve the equation x=\frac{9±3\sqrt{17}}{4} when ± is plus. Add 9 to 3\sqrt{17}.
x=\frac{9-3\sqrt{17}}{4}
Now solve the equation x=\frac{9±3\sqrt{17}}{4} when ± is minus. Subtract 3\sqrt{17} from 9.
x=\frac{3\sqrt{17}+9}{4} x=\frac{9-3\sqrt{17}}{4}
The equation is now solved.
2x^{2}-9x-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.
2x^{2}-9x-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
2x^{2}-9x=-\left(-9\right)
Subtracting -9 from itself leaves 0.
2x^{2}-9x=9
Subtract -9 from 0.
\frac{2x^{2}-9x}{2}=\frac{9}{2}
Divide both sides by 2.
x^{2}-\frac{9}{2}x=\frac{9}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\frac{9}{2}+\left(-\frac{9}{4}\right)^{2}
Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{9}{2}+\frac{81}{16}
Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{153}{16}
Add \frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{9}{4}\right)^{2}=\frac{153}{16}
Factor x^{2}-\frac{9}{2}x+\frac{81}{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{9}{4}\right)^{2}}=\sqrt{\frac{153}{16}}
Take the square root of both sides of the equation.
x-\frac{9}{4}=\frac{3\sqrt{17}}{4} x-\frac{9}{4}=-\frac{3\sqrt{17}}{4}
Simplify.
x=\frac{3\sqrt{17}+9}{4} x=\frac{9-3\sqrt{17}}{4}
Add \frac{9}{4} to both sides of the equation.
x ^ 2 -\frac{9}{2}x -\frac{9}{2} = 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 2
r + s = \frac{9}{2} rs = -\frac{9}{2}
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}{4} - u s = \frac{9}{4} + u
Two numbers r and s sum up to \frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{2} = \frac{9}{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{9}{4} - u) (\frac{9}{4} + u) = -\frac{9}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{2}
\frac{81}{16} - u^2 = -\frac{9}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{2}-\frac{81}{16} = -\frac{153}{16}
Simplify the expression by subtracting \frac{81}{16} on both sides
u^2 = \frac{153}{16} u = \pm\sqrt{\frac{153}{16}} = \pm \frac{\sqrt{153}}{4}
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}{4} - \frac{\sqrt{153}}{4} = -0.842 s = \frac{9}{4} + \frac{\sqrt{153}}{4} = 5.342
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.