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x^{2}-110x+55=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(-110\right)±\sqrt{\left(-110\right)^{2}-4\times 55}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -110 for b, and 55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-110\right)±\sqrt{12100-4\times 55}}{2}
Square -110.
x=\frac{-\left(-110\right)±\sqrt{12100-220}}{2}
Multiply -4 times 55.
x=\frac{-\left(-110\right)±\sqrt{11880}}{2}
Add 12100 to -220.
x=\frac{-\left(-110\right)±6\sqrt{330}}{2}
Take the square root of 11880.
x=\frac{110±6\sqrt{330}}{2}
The opposite of -110 is 110.
x=\frac{6\sqrt{330}+110}{2}
Now solve the equation x=\frac{110±6\sqrt{330}}{2} when ± is plus. Add 110 to 6\sqrt{330}.
x=3\sqrt{330}+55
Divide 110+6\sqrt{330} by 2.
x=\frac{110-6\sqrt{330}}{2}
Now solve the equation x=\frac{110±6\sqrt{330}}{2} when ± is minus. Subtract 6\sqrt{330} from 110.
x=55-3\sqrt{330}
Divide 110-6\sqrt{330} by 2.
x=3\sqrt{330}+55 x=55-3\sqrt{330}
The equation is now solved.
x^{2}-110x+55=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}-110x+55-55=-55
Subtract 55 from both sides of the equation.
x^{2}-110x=-55
Subtracting 55 from itself leaves 0.
x^{2}-110x+\left(-55\right)^{2}=-55+\left(-55\right)^{2}
Divide -110, the coefficient of the x term, by 2 to get -55. Then add the square of -55 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-110x+3025=-55+3025
Square -55.
x^{2}-110x+3025=2970
Add -55 to 3025.
\left(x-55\right)^{2}=2970
Factor x^{2}-110x+3025. 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-55\right)^{2}}=\sqrt{2970}
Take the square root of both sides of the equation.
x-55=3\sqrt{330} x-55=-3\sqrt{330}
Simplify.
x=3\sqrt{330}+55 x=55-3\sqrt{330}
Add 55 to both sides of the equation.
x ^ 2 -110x +55 = 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 = 110 rs = 55
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 = 55 - u s = 55 + u
Two numbers r and s sum up to 110 exactly when the average of the two numbers is \frac{1}{2}*110 = 55. 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>
(55 - u) (55 + u) = 55
To solve for unknown quantity u, substitute these in the product equation rs = 55
3025 - u^2 = 55
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 55-3025 = -2970
Simplify the expression by subtracting 3025 on both sides
u^2 = 2970 u = \pm\sqrt{2970} = \pm \sqrt{2970}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =55 - \sqrt{2970} = 0.502 s = 55 + \sqrt{2970} = 109.498
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.