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2v^{2}+17v+5=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.
v=\frac{-17±\sqrt{17^{2}-4\times 2\times 5}}{2\times 2}
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.
v=\frac{-17±\sqrt{289-4\times 2\times 5}}{2\times 2}
Square 17.
v=\frac{-17±\sqrt{289-8\times 5}}{2\times 2}
Multiply -4 times 2.
v=\frac{-17±\sqrt{289-40}}{2\times 2}
Multiply -8 times 5.
v=\frac{-17±\sqrt{249}}{2\times 2}
Add 289 to -40.
v=\frac{-17±\sqrt{249}}{4}
Multiply 2 times 2.
v=\frac{\sqrt{249}-17}{4}
Now solve the equation v=\frac{-17±\sqrt{249}}{4} when ± is plus. Add -17 to \sqrt{249}.
v=\frac{-\sqrt{249}-17}{4}
Now solve the equation v=\frac{-17±\sqrt{249}}{4} when ± is minus. Subtract \sqrt{249} from -17.
2v^{2}+17v+5=2\left(v-\frac{\sqrt{249}-17}{4}\right)\left(v-\frac{-\sqrt{249}-17}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-17+\sqrt{249}}{4} for x_{1} and \frac{-17-\sqrt{249}}{4} for x_{2}.
x ^ 2 +\frac{17}{2}x +\frac{5}{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{17}{2} rs = \frac{5}{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{17}{4} - u s = -\frac{17}{4} + u
Two numbers r and s sum up to -\frac{17}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{2} = -\frac{17}{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{17}{4} - u) (-\frac{17}{4} + u) = \frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}
\frac{289}{16} - u^2 = \frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{2}-\frac{289}{16} = -\frac{249}{16}
Simplify the expression by subtracting \frac{289}{16} on both sides
u^2 = \frac{249}{16} u = \pm\sqrt{\frac{249}{16}} = \pm \frac{\sqrt{249}}{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{17}{4} - \frac{\sqrt{249}}{4} = -8.195 s = -\frac{17}{4} + \frac{\sqrt{249}}{4} = -0.305
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.