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a+b=-5 ab=6\left(-4\right)=-24
Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-4. To find a and b, set up a system to be solved.
1,-24 2,-12 3,-8 4,-6
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.
1-24=-23 2-12=-10 3-8=-5 4-6=-2
Calculate the sum for each pair.
a=-8 b=3
The solution is the pair that gives sum -5.
\left(6y^{2}-8y\right)+\left(3y-4\right)
Rewrite 6y^{2}-5y-4 as \left(6y^{2}-8y\right)+\left(3y-4\right).
2y\left(3y-4\right)+3y-4
Factor out 2y in 6y^{2}-8y.
\left(3y-4\right)\left(2y+1\right)
Factor out common term 3y-4 by using distributive property.
6y^{2}-5y-4=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
y=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-4\right)}}{2\times 6}
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(-5\right)±\sqrt{25-4\times 6\left(-4\right)}}{2\times 6}
Square -5.
y=\frac{-\left(-5\right)±\sqrt{25-24\left(-4\right)}}{2\times 6}
Multiply -4 times 6.
y=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 6}
Multiply -24 times -4.
y=\frac{-\left(-5\right)±\sqrt{121}}{2\times 6}
Add 25 to 96.
y=\frac{-\left(-5\right)±11}{2\times 6}
Take the square root of 121.
y=\frac{5±11}{2\times 6}
The opposite of -5 is 5.
y=\frac{5±11}{12}
Multiply 2 times 6.
y=\frac{16}{12}
Now solve the equation y=\frac{5±11}{12} when ± is plus. Add 5 to 11.
y=\frac{4}{3}
Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.
y=-\frac{6}{12}
Now solve the equation y=\frac{5±11}{12} when ± is minus. Subtract 11 from 5.
y=-\frac{1}{2}
Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.
6y^{2}-5y-4=6\left(y-\frac{4}{3}\right)\left(y-\left(-\frac{1}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{1}{2} for x_{2}.
6y^{2}-5y-4=6\left(y-\frac{4}{3}\right)\left(y+\frac{1}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
6y^{2}-5y-4=6\times \frac{3y-4}{3}\left(y+\frac{1}{2}\right)
Subtract \frac{4}{3} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
6y^{2}-5y-4=6\times \frac{3y-4}{3}\times \frac{2y+1}{2}
Add \frac{1}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
6y^{2}-5y-4=6\times \frac{\left(3y-4\right)\left(2y+1\right)}{3\times 2}
Multiply \frac{3y-4}{3} times \frac{2y+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
6y^{2}-5y-4=6\times \frac{\left(3y-4\right)\left(2y+1\right)}{6}
Multiply 3 times 2.
6y^{2}-5y-4=\left(3y-4\right)\left(2y+1\right)
Cancel out 6, the greatest common factor in 6 and 6.
x ^ 2 -\frac{5}{6}x -\frac{2}{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 6
r + s = \frac{5}{6} rs = -\frac{2}{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{5}{12} - u s = \frac{5}{12} + u
Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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{5}{12} - u) (\frac{5}{12} + u) = -\frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}
\frac{25}{144} - u^2 = -\frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{3}-\frac{25}{144} = -\frac{121}{144}
Simplify the expression by subtracting \frac{25}{144} on both sides
u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{12} - \frac{11}{12} = -0.500 s = \frac{5}{12} + \frac{11}{12} = 1.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.