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4x^{2}-232x+2496=0
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(-232\right)±\sqrt{\left(-232\right)^{2}-4\times 4\times 2496}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -232 for b, and 2496 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-232\right)±\sqrt{53824-4\times 4\times 2496}}{2\times 4}
Square -232.
x=\frac{-\left(-232\right)±\sqrt{53824-16\times 2496}}{2\times 4}
Multiply -4 times 4.
x=\frac{-\left(-232\right)±\sqrt{53824-39936}}{2\times 4}
Multiply -16 times 2496.
x=\frac{-\left(-232\right)±\sqrt{13888}}{2\times 4}
Add 53824 to -39936.
x=\frac{-\left(-232\right)±8\sqrt{217}}{2\times 4}
Take the square root of 13888.
x=\frac{232±8\sqrt{217}}{2\times 4}
The opposite of -232 is 232.
x=\frac{232±8\sqrt{217}}{8}
Multiply 2 times 4.
x=\frac{8\sqrt{217}+232}{8}
Now solve the equation x=\frac{232±8\sqrt{217}}{8} when ± is plus. Add 232 to 8\sqrt{217}.
x=\sqrt{217}+29
Divide 232+8\sqrt{217} by 8.
x=\frac{232-8\sqrt{217}}{8}
Now solve the equation x=\frac{232±8\sqrt{217}}{8} when ± is minus. Subtract 8\sqrt{217} from 232.
x=29-\sqrt{217}
Divide 232-8\sqrt{217} by 8.
x=\sqrt{217}+29 x=29-\sqrt{217}
The equation is now solved.
4x^{2}-232x+2496=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
4x^{2}-232x+2496-2496=-2496
Subtract 2496 from both sides of the equation.
4x^{2}-232x=-2496
Subtracting 2496 from itself leaves 0.
\frac{4x^{2}-232x}{4}=-\frac{2496}{4}
Divide both sides by 4.
x^{2}+\left(-\frac{232}{4}\right)x=-\frac{2496}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}-58x=-\frac{2496}{4}
Divide -232 by 4.
x^{2}-58x=-624
Divide -2496 by 4.
x^{2}-58x+\left(-29\right)^{2}=-624+\left(-29\right)^{2}
Divide -58, the coefficient of the x term, by 2 to get -29. Then add the square of -29 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-58x+841=-624+841
Square -29.
x^{2}-58x+841=217
Add -624 to 841.
\left(x-29\right)^{2}=217
Factor x^{2}-58x+841. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-29\right)^{2}}=\sqrt{217}
Take the square root of both sides of the equation.
x-29=\sqrt{217} x-29=-\sqrt{217}
Simplify.
x=\sqrt{217}+29 x=29-\sqrt{217}
Add 29 to both sides of the equation.
x ^ 2 -58x +624 = 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 4
r + s = 58 rs = 624
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 = 29 - u s = 29 + u
Two numbers r and s sum up to 58 exactly when the average of the two numbers is \frac{1}{2}*58 = 29. 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>
(29 - u) (29 + u) = 624
To solve for unknown quantity u, substitute these in the product equation rs = 624
841 - u^2 = 624
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 624-841 = -217
Simplify the expression by subtracting 841 on both sides
u^2 = 217 u = \pm\sqrt{217} = \pm \sqrt{217}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =29 - \sqrt{217} = 14.269 s = 29 + \sqrt{217} = 43.731
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.