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5m^{2}-13m-15=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.
m=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 5\left(-15\right)}}{2\times 5}
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.
m=\frac{-\left(-13\right)±\sqrt{169-4\times 5\left(-15\right)}}{2\times 5}
Square -13.
m=\frac{-\left(-13\right)±\sqrt{169-20\left(-15\right)}}{2\times 5}
Multiply -4 times 5.
m=\frac{-\left(-13\right)±\sqrt{169+300}}{2\times 5}
Multiply -20 times -15.
m=\frac{-\left(-13\right)±\sqrt{469}}{2\times 5}
Add 169 to 300.
m=\frac{13±\sqrt{469}}{2\times 5}
The opposite of -13 is 13.
m=\frac{13±\sqrt{469}}{10}
Multiply 2 times 5.
m=\frac{\sqrt{469}+13}{10}
Now solve the equation m=\frac{13±\sqrt{469}}{10} when ± is plus. Add 13 to \sqrt{469}.
m=\frac{13-\sqrt{469}}{10}
Now solve the equation m=\frac{13±\sqrt{469}}{10} when ± is minus. Subtract \sqrt{469} from 13.
5m^{2}-13m-15=5\left(m-\frac{\sqrt{469}+13}{10}\right)\left(m-\frac{13-\sqrt{469}}{10}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13+\sqrt{469}}{10} for x_{1} and \frac{13-\sqrt{469}}{10} for x_{2}.
x ^ 2 -\frac{13}{5}x -3 = 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 5
r + s = \frac{13}{5} rs = -3
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{13}{10} - u s = \frac{13}{10} + u
Two numbers r and s sum up to \frac{13}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{5} = \frac{13}{10}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{13}{10} - u) (\frac{13}{10} + u) = -3
To solve for unknown quantity u, substitute these in the product equation rs = -3
\frac{169}{100} - u^2 = -3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -3-\frac{169}{100} = -\frac{469}{100}
Simplify the expression by subtracting \frac{169}{100} on both sides
u^2 = \frac{469}{100} u = \pm\sqrt{\frac{469}{100}} = \pm \frac{\sqrt{469}}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{13}{10} - \frac{\sqrt{469}}{10} = -0.866 s = \frac{13}{10} + \frac{\sqrt{469}}{10} = 3.466
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.