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4y^{2}-y-18=0
Subtract 18 from both sides.
a+b=-1 ab=4\left(-18\right)=-72
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4y^{2}+ay+by-18. To find a and b, set up a system to be solved.
1,-72 2,-36 3,-24 4,-18 6,-12 8,-9
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 -72.
1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1
Calculate the sum for each pair.
a=-9 b=8
The solution is the pair that gives sum -1.
\left(4y^{2}-9y\right)+\left(8y-18\right)
Rewrite 4y^{2}-y-18 as \left(4y^{2}-9y\right)+\left(8y-18\right).
y\left(4y-9\right)+2\left(4y-9\right)
Factor out y in the first and 2 in the second group.
\left(4y-9\right)\left(y+2\right)
Factor out common term 4y-9 by using distributive property.
y=\frac{9}{4} y=-2
To find equation solutions, solve 4y-9=0 and y+2=0.
4y^{2}-y=18
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.
4y^{2}-y-18=18-18
Subtract 18 from both sides of the equation.
4y^{2}-y-18=0
Subtracting 18 from itself leaves 0.
y=\frac{-\left(-1\right)±\sqrt{1-4\times 4\left(-18\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -1 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-1\right)±\sqrt{1-16\left(-18\right)}}{2\times 4}
Multiply -4 times 4.
y=\frac{-\left(-1\right)±\sqrt{1+288}}{2\times 4}
Multiply -16 times -18.
y=\frac{-\left(-1\right)±\sqrt{289}}{2\times 4}
Add 1 to 288.
y=\frac{-\left(-1\right)±17}{2\times 4}
Take the square root of 289.
y=\frac{1±17}{2\times 4}
The opposite of -1 is 1.
y=\frac{1±17}{8}
Multiply 2 times 4.
y=\frac{18}{8}
Now solve the equation y=\frac{1±17}{8} when ± is plus. Add 1 to 17.
y=\frac{9}{4}
Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.
y=-\frac{16}{8}
Now solve the equation y=\frac{1±17}{8} when ± is minus. Subtract 17 from 1.
y=-2
Divide -16 by 8.
y=\frac{9}{4} y=-2
The equation is now solved.
4y^{2}-y=18
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.
\frac{4y^{2}-y}{4}=\frac{18}{4}
Divide both sides by 4.
y^{2}-\frac{1}{4}y=\frac{18}{4}
Dividing by 4 undoes the multiplication by 4.
y^{2}-\frac{1}{4}y=\frac{9}{2}
Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.
y^{2}-\frac{1}{4}y+\left(-\frac{1}{8}\right)^{2}=\frac{9}{2}+\left(-\frac{1}{8}\right)^{2}
Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{1}{4}y+\frac{1}{64}=\frac{9}{2}+\frac{1}{64}
Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{1}{4}y+\frac{1}{64}=\frac{289}{64}
Add \frac{9}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{1}{8}\right)^{2}=\frac{289}{64}
Factor y^{2}-\frac{1}{4}y+\frac{1}{64}. 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{1}{8}\right)^{2}}=\sqrt{\frac{289}{64}}
Take the square root of both sides of the equation.
y-\frac{1}{8}=\frac{17}{8} y-\frac{1}{8}=-\frac{17}{8}
Simplify.
y=\frac{9}{4} y=-2
Add \frac{1}{8} to both sides of the equation.