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65y^{2}-23y-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.
y=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 65\left(-10\right)}}{2\times 65}
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.
y=\frac{-\left(-23\right)±\sqrt{529-4\times 65\left(-10\right)}}{2\times 65}
Square -23.
y=\frac{-\left(-23\right)±\sqrt{529-260\left(-10\right)}}{2\times 65}
Multiply -4 times 65.
y=\frac{-\left(-23\right)±\sqrt{529+2600}}{2\times 65}
Multiply -260 times -10.
y=\frac{-\left(-23\right)±\sqrt{3129}}{2\times 65}
Add 529 to 2600.
y=\frac{23±\sqrt{3129}}{2\times 65}
The opposite of -23 is 23.
y=\frac{23±\sqrt{3129}}{130}
Multiply 2 times 65.
y=\frac{\sqrt{3129}+23}{130}
Now solve the equation y=\frac{23±\sqrt{3129}}{130} when ± is plus. Add 23 to \sqrt{3129}.
y=\frac{23-\sqrt{3129}}{130}
Now solve the equation y=\frac{23±\sqrt{3129}}{130} when ± is minus. Subtract \sqrt{3129} from 23.
65y^{2}-23y-10=65\left(y-\frac{\sqrt{3129}+23}{130}\right)\left(y-\frac{23-\sqrt{3129}}{130}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{23+\sqrt{3129}}{130} for x_{1} and \frac{23-\sqrt{3129}}{130} for x_{2}.
x ^ 2 -\frac{23}{65}x -\frac{2}{13} = 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 65
r + s = \frac{23}{65} rs = -\frac{2}{13}
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{23}{130} - u s = \frac{23}{130} + u
Two numbers r and s sum up to \frac{23}{65} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{65} = \frac{23}{130}. 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{23}{130} - u) (\frac{23}{130} + u) = -\frac{2}{13}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{13}
\frac{529}{16900} - u^2 = -\frac{2}{13}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{13}-\frac{529}{16900} = -\frac{3129}{16900}
Simplify the expression by subtracting \frac{529}{16900} on both sides
u^2 = \frac{3129}{16900} u = \pm\sqrt{\frac{3129}{16900}} = \pm \frac{\sqrt{3129}}{130}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{23}{130} - \frac{\sqrt{3129}}{130} = -0.253 s = \frac{23}{130} + \frac{\sqrt{3129}}{130} = 0.607
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.