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2x^{2}-30x-58=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{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 2\left(-58\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{-\left(-30\right)±\sqrt{900-4\times 2\left(-58\right)}}{2\times 2}
Square -30.
x=\frac{-\left(-30\right)±\sqrt{900-8\left(-58\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-30\right)±\sqrt{900+464}}{2\times 2}
Multiply -8 times -58.
x=\frac{-\left(-30\right)±\sqrt{1364}}{2\times 2}
Add 900 to 464.
x=\frac{-\left(-30\right)±2\sqrt{341}}{2\times 2}
Take the square root of 1364.
x=\frac{30±2\sqrt{341}}{2\times 2}
The opposite of -30 is 30.
x=\frac{30±2\sqrt{341}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{341}+30}{4}
Now solve the equation x=\frac{30±2\sqrt{341}}{4} when ± is plus. Add 30 to 2\sqrt{341}.
x=\frac{\sqrt{341}+15}{2}
Divide 30+2\sqrt{341} by 4.
x=\frac{30-2\sqrt{341}}{4}
Now solve the equation x=\frac{30±2\sqrt{341}}{4} when ± is minus. Subtract 2\sqrt{341} from 30.
x=\frac{15-\sqrt{341}}{2}
Divide 30-2\sqrt{341} by 4.
2x^{2}-30x-58=2\left(x-\frac{\sqrt{341}+15}{2}\right)\left(x-\frac{15-\sqrt{341}}{2}\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{341}}{2} for x_{1} and \frac{15-\sqrt{341}}{2} for x_{2}.
x ^ 2 -15x -29 = 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 = 15 rs = -29
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}{2} - u s = \frac{15}{2} + u
Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{2}. 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}{2} - u) (\frac{15}{2} + u) = -29
To solve for unknown quantity u, substitute these in the product equation rs = -29
\frac{225}{4} - u^2 = -29
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -29-\frac{225}{4} = -\frac{341}{4}
Simplify the expression by subtracting \frac{225}{4} on both sides
u^2 = \frac{341}{4} u = \pm\sqrt{\frac{341}{4}} = \pm \frac{\sqrt{341}}{2}
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}{2} - \frac{\sqrt{341}}{2} = -1.733 s = \frac{15}{2} + \frac{\sqrt{341}}{2} = 16.733
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.