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54x^{2}-7x-3=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 54\left(-3\right)}}{2\times 54}
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(-7\right)±\sqrt{49-4\times 54\left(-3\right)}}{2\times 54}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-216\left(-3\right)}}{2\times 54}
Multiply -4 times 54.
x=\frac{-\left(-7\right)±\sqrt{49+648}}{2\times 54}
Multiply -216 times -3.
x=\frac{-\left(-7\right)±\sqrt{697}}{2\times 54}
Add 49 to 648.
x=\frac{7±\sqrt{697}}{2\times 54}
The opposite of -7 is 7.
x=\frac{7±\sqrt{697}}{108}
Multiply 2 times 54.
x=\frac{\sqrt{697}+7}{108}
Now solve the equation x=\frac{7±\sqrt{697}}{108} when ± is plus. Add 7 to \sqrt{697}.
x=\frac{7-\sqrt{697}}{108}
Now solve the equation x=\frac{7±\sqrt{697}}{108} when ± is minus. Subtract \sqrt{697} from 7.
54x^{2}-7x-3=54\left(x-\frac{\sqrt{697}+7}{108}\right)\left(x-\frac{7-\sqrt{697}}{108}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7+\sqrt{697}}{108} for x_{1} and \frac{7-\sqrt{697}}{108} for x_{2}.
x ^ 2 -\frac{7}{54}x -\frac{1}{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 54
r + s = \frac{7}{54} rs = -\frac{1}{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{7}{108} - u s = \frac{7}{108} + u
Two numbers r and s sum up to \frac{7}{54} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{54} = \frac{7}{108}. 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{7}{108} - u) (\frac{7}{108} + u) = -\frac{1}{18}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{18}
\frac{49}{11664} - u^2 = -\frac{1}{18}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{18}-\frac{49}{11664} = -\frac{697}{11664}
Simplify the expression by subtracting \frac{49}{11664} on both sides
u^2 = \frac{697}{11664} u = \pm\sqrt{\frac{697}{11664}} = \pm \frac{\sqrt{697}}{108}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{108} - \frac{\sqrt{697}}{108} = -0.180 s = \frac{7}{108} + \frac{\sqrt{697}}{108} = 0.309
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.