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2x^{2}+15x-112=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.
x=\frac{-15±\sqrt{15^{2}-4\times 2\left(-112\right)}}{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.
x=\frac{-15±\sqrt{225-4\times 2\left(-112\right)}}{2\times 2}
Square 15.
x=\frac{-15±\sqrt{225-8\left(-112\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-15±\sqrt{225+896}}{2\times 2}
Multiply -8 times -112.
x=\frac{-15±\sqrt{1121}}{2\times 2}
Add 225 to 896.
x=\frac{-15±\sqrt{1121}}{4}
Multiply 2 times 2.
x=\frac{\sqrt{1121}-15}{4}
Now solve the equation x=\frac{-15±\sqrt{1121}}{4} when ± is plus. Add -15 to \sqrt{1121}.
x=\frac{-\sqrt{1121}-15}{4}
Now solve the equation x=\frac{-15±\sqrt{1121}}{4} when ± is minus. Subtract \sqrt{1121} from -15.
2x^{2}+15x-112=2\left(x-\frac{\sqrt{1121}-15}{4}\right)\left(x-\frac{-\sqrt{1121}-15}{4}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-15+\sqrt{1121}}{4} for x_{1} and \frac{-15-\sqrt{1121}}{4} for x_{2}.
x ^ 2 +\frac{15}{2}x -56 = 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{15}{2} rs = -56
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{15}{4} - u s = -\frac{15}{4} + u
Two numbers r and s sum up to -\frac{15}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{15}{2} = -\frac{15}{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{15}{4} - u) (-\frac{15}{4} + u) = -56
To solve for unknown quantity u, substitute these in the product equation rs = -56
\frac{225}{16} - u^2 = -56
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -56-\frac{225}{16} = -\frac{1121}{16}
Simplify the expression by subtracting \frac{225}{16} on both sides
u^2 = \frac{1121}{16} u = \pm\sqrt{\frac{1121}{16}} = \pm \frac{\sqrt{1121}}{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{15}{4} - \frac{\sqrt{1121}}{4} = -12.120 s = -\frac{15}{4} + \frac{\sqrt{1121}}{4} = 4.620
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.