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18x^{2}+24x+7=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{-24±\sqrt{24^{2}-4\times 18\times 7}}{2\times 18}
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{-24±\sqrt{576-4\times 18\times 7}}{2\times 18}
Square 24.
x=\frac{-24±\sqrt{576-72\times 7}}{2\times 18}
Multiply -4 times 18.
x=\frac{-24±\sqrt{576-504}}{2\times 18}
Multiply -72 times 7.
x=\frac{-24±\sqrt{72}}{2\times 18}
Add 576 to -504.
x=\frac{-24±6\sqrt{2}}{2\times 18}
Take the square root of 72.
x=\frac{-24±6\sqrt{2}}{36}
Multiply 2 times 18.
x=\frac{6\sqrt{2}-24}{36}
Now solve the equation x=\frac{-24±6\sqrt{2}}{36} when ± is plus. Add -24 to 6\sqrt{2}.
x=\frac{\sqrt{2}}{6}-\frac{2}{3}
Divide -24+6\sqrt{2} by 36.
x=\frac{-6\sqrt{2}-24}{36}
Now solve the equation x=\frac{-24±6\sqrt{2}}{36} when ± is minus. Subtract 6\sqrt{2} from -24.
x=-\frac{\sqrt{2}}{6}-\frac{2}{3}
Divide -24-6\sqrt{2} by 36.
18x^{2}+24x+7=18\left(x-\left(\frac{\sqrt{2}}{6}-\frac{2}{3}\right)\right)\left(x-\left(-\frac{\sqrt{2}}{6}-\frac{2}{3}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{2}{3}+\frac{\sqrt{2}}{6} for x_{1} and -\frac{2}{3}-\frac{\sqrt{2}}{6} for x_{2}.
x ^ 2 +\frac{4}{3}x +\frac{7}{18} = 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 18
r + s = -\frac{4}{3} rs = \frac{7}{18}
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{2}{3} - u s = -\frac{2}{3} + u
Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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{2}{3} - u) (-\frac{2}{3} + u) = \frac{7}{18}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{18}
\frac{4}{9} - u^2 = \frac{7}{18}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{7}{18}-\frac{4}{9} = -\frac{1}{18}
Simplify the expression by subtracting \frac{4}{9} on both sides
u^2 = \frac{1}{18} u = \pm\sqrt{\frac{1}{18}} = \pm \frac{1}{\sqrt{18}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{2}{3} - \frac{1}{\sqrt{18}} = -0.902 s = -\frac{2}{3} + \frac{1}{\sqrt{18}} = -0.431
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.