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x^{2}-56x+544=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(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 544}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -56 for b, and 544 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-56\right)±\sqrt{3136-4\times 544}}{2}
Square -56.
x=\frac{-\left(-56\right)±\sqrt{3136-2176}}{2}
Multiply -4 times 544.
x=\frac{-\left(-56\right)±\sqrt{960}}{2}
Add 3136 to -2176.
x=\frac{-\left(-56\right)±8\sqrt{15}}{2}
Take the square root of 960.
x=\frac{56±8\sqrt{15}}{2}
The opposite of -56 is 56.
x=\frac{8\sqrt{15}+56}{2}
Now solve the equation x=\frac{56±8\sqrt{15}}{2} when ± is plus. Add 56 to 8\sqrt{15}.
x=4\sqrt{15}+28
Divide 56+8\sqrt{15} by 2.
x=\frac{56-8\sqrt{15}}{2}
Now solve the equation x=\frac{56±8\sqrt{15}}{2} when ± is minus. Subtract 8\sqrt{15} from 56.
x=28-4\sqrt{15}
Divide 56-8\sqrt{15} by 2.
x=4\sqrt{15}+28 x=28-4\sqrt{15}
The equation is now solved.
x^{2}-56x+544=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.
x^{2}-56x+544-544=-544
Subtract 544 from both sides of the equation.
x^{2}-56x=-544
Subtracting 544 from itself leaves 0.
x^{2}-56x+\left(-28\right)^{2}=-544+\left(-28\right)^{2}
Divide -56, the coefficient of the x term, by 2 to get -28. Then add the square of -28 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-56x+784=-544+784
Square -28.
x^{2}-56x+784=240
Add -544 to 784.
\left(x-28\right)^{2}=240
Factor x^{2}-56x+784. 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-28\right)^{2}}=\sqrt{240}
Take the square root of both sides of the equation.
x-28=4\sqrt{15} x-28=-4\sqrt{15}
Simplify.
x=4\sqrt{15}+28 x=28-4\sqrt{15}
Add 28 to both sides of the equation.
x ^ 2 -56x +544 = 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.
r + s = 56 rs = 544
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 = 28 - u s = 28 + u
Two numbers r and s sum up to 56 exactly when the average of the two numbers is \frac{1}{2}*56 = 28. 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>
(28 - u) (28 + u) = 544
To solve for unknown quantity u, substitute these in the product equation rs = 544
784 - u^2 = 544
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 544-784 = -240
Simplify the expression by subtracting 784 on both sides
u^2 = 240 u = \pm\sqrt{240} = \pm \sqrt{240}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =28 - \sqrt{240} = 12.508 s = 28 + \sqrt{240} = 43.492
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.