Solve for a
a = \frac{\sqrt{2785} + 55}{6} \approx 17.962183201
a=\frac{55-\sqrt{2785}}{6}\approx 0.371150132
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3a^{2}-55a+20=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.
a=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\times 3\times 20}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -55 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-55\right)±\sqrt{3025-4\times 3\times 20}}{2\times 3}
Square -55.
a=\frac{-\left(-55\right)±\sqrt{3025-12\times 20}}{2\times 3}
Multiply -4 times 3.
a=\frac{-\left(-55\right)±\sqrt{3025-240}}{2\times 3}
Multiply -12 times 20.
a=\frac{-\left(-55\right)±\sqrt{2785}}{2\times 3}
Add 3025 to -240.
a=\frac{55±\sqrt{2785}}{2\times 3}
The opposite of -55 is 55.
a=\frac{55±\sqrt{2785}}{6}
Multiply 2 times 3.
a=\frac{\sqrt{2785}+55}{6}
Now solve the equation a=\frac{55±\sqrt{2785}}{6} when ± is plus. Add 55 to \sqrt{2785}.
a=\frac{55-\sqrt{2785}}{6}
Now solve the equation a=\frac{55±\sqrt{2785}}{6} when ± is minus. Subtract \sqrt{2785} from 55.
a=\frac{\sqrt{2785}+55}{6} a=\frac{55-\sqrt{2785}}{6}
The equation is now solved.
3a^{2}-55a+20=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.
3a^{2}-55a+20-20=-20
Subtract 20 from both sides of the equation.
3a^{2}-55a=-20
Subtracting 20 from itself leaves 0.
\frac{3a^{2}-55a}{3}=-\frac{20}{3}
Divide both sides by 3.
a^{2}-\frac{55}{3}a=-\frac{20}{3}
Dividing by 3 undoes the multiplication by 3.
a^{2}-\frac{55}{3}a+\left(-\frac{55}{6}\right)^{2}=-\frac{20}{3}+\left(-\frac{55}{6}\right)^{2}
Divide -\frac{55}{3}, the coefficient of the x term, by 2 to get -\frac{55}{6}. Then add the square of -\frac{55}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{55}{3}a+\frac{3025}{36}=-\frac{20}{3}+\frac{3025}{36}
Square -\frac{55}{6} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{55}{3}a+\frac{3025}{36}=\frac{2785}{36}
Add -\frac{20}{3} to \frac{3025}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{55}{6}\right)^{2}=\frac{2785}{36}
Factor a^{2}-\frac{55}{3}a+\frac{3025}{36}. 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(a-\frac{55}{6}\right)^{2}}=\sqrt{\frac{2785}{36}}
Take the square root of both sides of the equation.
a-\frac{55}{6}=\frac{\sqrt{2785}}{6} a-\frac{55}{6}=-\frac{\sqrt{2785}}{6}
Simplify.
a=\frac{\sqrt{2785}+55}{6} a=\frac{55-\sqrt{2785}}{6}
Add \frac{55}{6} to both sides of the equation.
x ^ 2 -\frac{55}{3}x +\frac{20}{3} = 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 3
r + s = \frac{55}{3} rs = \frac{20}{3}
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{55}{6} - u s = \frac{55}{6} + u
Two numbers r and s sum up to \frac{55}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{55}{3} = \frac{55}{6}. 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{55}{6} - u) (\frac{55}{6} + u) = \frac{20}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{3}
\frac{3025}{36} - u^2 = \frac{20}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{20}{3}-\frac{3025}{36} = -\frac{2785}{36}
Simplify the expression by subtracting \frac{3025}{36} on both sides
u^2 = \frac{2785}{36} u = \pm\sqrt{\frac{2785}{36}} = \pm \frac{\sqrt{2785}}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{55}{6} - \frac{\sqrt{2785}}{6} = 0.371 s = \frac{55}{6} + \frac{\sqrt{2785}}{6} = 17.962
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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