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88y^{2}-583y+330=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.
y=\frac{-\left(-583\right)±\sqrt{\left(-583\right)^{2}-4\times 88\times 330}}{2\times 88}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 88 for a, -583 for b, and 330 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-583\right)±\sqrt{339889-4\times 88\times 330}}{2\times 88}
Square -583.
y=\frac{-\left(-583\right)±\sqrt{339889-352\times 330}}{2\times 88}
Multiply -4 times 88.
y=\frac{-\left(-583\right)±\sqrt{339889-116160}}{2\times 88}
Multiply -352 times 330.
y=\frac{-\left(-583\right)±\sqrt{223729}}{2\times 88}
Add 339889 to -116160.
y=\frac{-\left(-583\right)±473}{2\times 88}
Take the square root of 223729.
y=\frac{583±473}{2\times 88}
The opposite of -583 is 583.
y=\frac{583±473}{176}
Multiply 2 times 88.
y=\frac{1056}{176}
Now solve the equation y=\frac{583±473}{176} when ± is plus. Add 583 to 473.
y=6
Divide 1056 by 176.
y=\frac{110}{176}
Now solve the equation y=\frac{583±473}{176} when ± is minus. Subtract 473 from 583.
y=\frac{5}{8}
Reduce the fraction \frac{110}{176} to lowest terms by extracting and canceling out 22.
y=6 y=\frac{5}{8}
The equation is now solved.
88y^{2}-583y+330=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.
88y^{2}-583y+330-330=-330
Subtract 330 from both sides of the equation.
88y^{2}-583y=-330
Subtracting 330 from itself leaves 0.
\frac{88y^{2}-583y}{88}=-\frac{330}{88}
Divide both sides by 88.
y^{2}+\left(-\frac{583}{88}\right)y=-\frac{330}{88}
Dividing by 88 undoes the multiplication by 88.
y^{2}-\frac{53}{8}y=-\frac{330}{88}
Reduce the fraction \frac{-583}{88} to lowest terms by extracting and canceling out 11.
y^{2}-\frac{53}{8}y=-\frac{15}{4}
Reduce the fraction \frac{-330}{88} to lowest terms by extracting and canceling out 22.
y^{2}-\frac{53}{8}y+\left(-\frac{53}{16}\right)^{2}=-\frac{15}{4}+\left(-\frac{53}{16}\right)^{2}
Divide -\frac{53}{8}, the coefficient of the x term, by 2 to get -\frac{53}{16}. Then add the square of -\frac{53}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{53}{8}y+\frac{2809}{256}=-\frac{15}{4}+\frac{2809}{256}
Square -\frac{53}{16} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{53}{8}y+\frac{2809}{256}=\frac{1849}{256}
Add -\frac{15}{4} to \frac{2809}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{53}{16}\right)^{2}=\frac{1849}{256}
Factor y^{2}-\frac{53}{8}y+\frac{2809}{256}. 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(y-\frac{53}{16}\right)^{2}}=\sqrt{\frac{1849}{256}}
Take the square root of both sides of the equation.
y-\frac{53}{16}=\frac{43}{16} y-\frac{53}{16}=-\frac{43}{16}
Simplify.
y=6 y=\frac{5}{8}
Add \frac{53}{16} to both sides of the equation.
x ^ 2 -\frac{53}{8}x +\frac{15}{4} = 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 88
r + s = \frac{53}{8} rs = \frac{15}{4}
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{53}{16} - u s = \frac{53}{16} + u
Two numbers r and s sum up to \frac{53}{8} exactly when the average of the two numbers is \frac{1}{2}*\frac{53}{8} = \frac{53}{16}. 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{53}{16} - u) (\frac{53}{16} + u) = \frac{15}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{4}
\frac{2809}{256} - u^2 = \frac{15}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{15}{4}-\frac{2809}{256} = -\frac{1849}{256}
Simplify the expression by subtracting \frac{2809}{256} on both sides
u^2 = \frac{1849}{256} u = \pm\sqrt{\frac{1849}{256}} = \pm \frac{43}{16}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{53}{16} - \frac{43}{16} = 0.625 s = \frac{53}{16} + \frac{43}{16} = 6
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.