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-2u^{2}+33u+10=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
u=\frac{-33±\sqrt{33^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}
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.
u=\frac{-33±\sqrt{1089-4\left(-2\right)\times 10}}{2\left(-2\right)}
Square 33.
u=\frac{-33±\sqrt{1089+8\times 10}}{2\left(-2\right)}
Multiply -4 times -2.
u=\frac{-33±\sqrt{1089+80}}{2\left(-2\right)}
Multiply 8 times 10.
u=\frac{-33±\sqrt{1169}}{2\left(-2\right)}
Add 1089 to 80.
u=\frac{-33±\sqrt{1169}}{-4}
Multiply 2 times -2.
u=\frac{\sqrt{1169}-33}{-4}
Now solve the equation u=\frac{-33±\sqrt{1169}}{-4} when ± is plus. Add -33 to \sqrt{1169}.
u=\frac{33-\sqrt{1169}}{4}
Divide -33+\sqrt{1169} by -4.
u=\frac{-\sqrt{1169}-33}{-4}
Now solve the equation u=\frac{-33±\sqrt{1169}}{-4} when ± is minus. Subtract \sqrt{1169} from -33.
u=\frac{\sqrt{1169}+33}{4}
Divide -33-\sqrt{1169} by -4.
-2u^{2}+33u+10=-2\left(u-\frac{33-\sqrt{1169}}{4}\right)\left(u-\frac{\sqrt{1169}+33}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{33-\sqrt{1169}}{4} for x_{1} and \frac{33+\sqrt{1169}}{4} for x_{2}.
x ^ 2 -\frac{33}{2}x -5 = 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{33}{2} rs = -5
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{33}{4} - u s = \frac{33}{4} + u
Two numbers r and s sum up to \frac{33}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{33}{2} = \frac{33}{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{33}{4} - u) (\frac{33}{4} + u) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
\frac{1089}{16} - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-\frac{1089}{16} = -\frac{1169}{16}
Simplify the expression by subtracting \frac{1089}{16} on both sides
u^2 = \frac{1169}{16} u = \pm\sqrt{\frac{1169}{16}} = \pm \frac{\sqrt{1169}}{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{33}{4} - \frac{\sqrt{1169}}{4} = -0.298 s = \frac{33}{4} + \frac{\sqrt{1169}}{4} = 16.798
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.