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10b^{2}-124b+144=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.
b=\frac{-\left(-124\right)±\sqrt{\left(-124\right)^{2}-4\times 10\times 144}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -124 for b, and 144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
b=\frac{-\left(-124\right)±\sqrt{15376-4\times 10\times 144}}{2\times 10}
Square -124.
b=\frac{-\left(-124\right)±\sqrt{15376-40\times 144}}{2\times 10}
Multiply -4 times 10.
b=\frac{-\left(-124\right)±\sqrt{15376-5760}}{2\times 10}
Multiply -40 times 144.
b=\frac{-\left(-124\right)±\sqrt{9616}}{2\times 10}
Add 15376 to -5760.
b=\frac{-\left(-124\right)±4\sqrt{601}}{2\times 10}
Take the square root of 9616.
b=\frac{124±4\sqrt{601}}{2\times 10}
The opposite of -124 is 124.
b=\frac{124±4\sqrt{601}}{20}
Multiply 2 times 10.
b=\frac{4\sqrt{601}+124}{20}
Now solve the equation b=\frac{124±4\sqrt{601}}{20} when ± is plus. Add 124 to 4\sqrt{601}.
b=\frac{\sqrt{601}+31}{5}
Divide 124+4\sqrt{601} by 20.
b=\frac{124-4\sqrt{601}}{20}
Now solve the equation b=\frac{124±4\sqrt{601}}{20} when ± is minus. Subtract 4\sqrt{601} from 124.
b=\frac{31-\sqrt{601}}{5}
Divide 124-4\sqrt{601} by 20.
b=\frac{\sqrt{601}+31}{5} b=\frac{31-\sqrt{601}}{5}
The equation is now solved.
10b^{2}-124b+144=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.
10b^{2}-124b+144-144=-144
Subtract 144 from both sides of the equation.
10b^{2}-124b=-144
Subtracting 144 from itself leaves 0.
\frac{10b^{2}-124b}{10}=-\frac{144}{10}
Divide both sides by 10.
b^{2}+\left(-\frac{124}{10}\right)b=-\frac{144}{10}
Dividing by 10 undoes the multiplication by 10.
b^{2}-\frac{62}{5}b=-\frac{144}{10}
Reduce the fraction \frac{-124}{10} to lowest terms by extracting and canceling out 2.
b^{2}-\frac{62}{5}b=-\frac{72}{5}
Reduce the fraction \frac{-144}{10} to lowest terms by extracting and canceling out 2.
b^{2}-\frac{62}{5}b+\left(-\frac{31}{5}\right)^{2}=-\frac{72}{5}+\left(-\frac{31}{5}\right)^{2}
Divide -\frac{62}{5}, the coefficient of the x term, by 2 to get -\frac{31}{5}. Then add the square of -\frac{31}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
b^{2}-\frac{62}{5}b+\frac{961}{25}=-\frac{72}{5}+\frac{961}{25}
Square -\frac{31}{5} by squaring both the numerator and the denominator of the fraction.
b^{2}-\frac{62}{5}b+\frac{961}{25}=\frac{601}{25}
Add -\frac{72}{5} to \frac{961}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(b-\frac{31}{5}\right)^{2}=\frac{601}{25}
Factor b^{2}-\frac{62}{5}b+\frac{961}{25}. 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(b-\frac{31}{5}\right)^{2}}=\sqrt{\frac{601}{25}}
Take the square root of both sides of the equation.
b-\frac{31}{5}=\frac{\sqrt{601}}{5} b-\frac{31}{5}=-\frac{\sqrt{601}}{5}
Simplify.
b=\frac{\sqrt{601}+31}{5} b=\frac{31-\sqrt{601}}{5}
Add \frac{31}{5} to both sides of the equation.
x ^ 2 -\frac{62}{5}x +\frac{72}{5} = 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 10
r + s = \frac{62}{5} rs = \frac{72}{5}
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{31}{5} - u s = \frac{31}{5} + u
Two numbers r and s sum up to \frac{62}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{62}{5} = \frac{31}{5}. 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{31}{5} - u) (\frac{31}{5} + u) = \frac{72}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{72}{5}
\frac{961}{25} - u^2 = \frac{72}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{72}{5}-\frac{961}{25} = -\frac{601}{25}
Simplify the expression by subtracting \frac{961}{25} on both sides
u^2 = \frac{601}{25} u = \pm\sqrt{\frac{601}{25}} = \pm \frac{\sqrt{601}}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{31}{5} - \frac{\sqrt{601}}{5} = 1.297 s = \frac{31}{5} + \frac{\sqrt{601}}{5} = 11.103
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.