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4x^{2}+x-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 4x^{2}+ax+bx-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 positive, the positive number has greater absolute value than the negative. 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=-8 b=9
The solution is the pair that gives sum 1.
\left(4x^{2}-8x\right)+\left(9x-18\right)
Rewrite 4x^{2}+x-18 as \left(4x^{2}-8x\right)+\left(9x-18\right).
4x\left(x-2\right)+9\left(x-2\right)
Factor out 4x in the first and 9 in the second group.
\left(x-2\right)\left(4x+9\right)
Factor out common term x-2 by using distributive property.
x=2 x=-\frac{9}{4}
To find equation solutions, solve x-2=0 and 4x+9=0.
4x^{2}+x=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.
4x^{2}+x-18=18-18
Subtract 18 from both sides of the equation.
4x^{2}+x-18=0
Subtracting 18 from itself leaves 0.
x=\frac{-1±\sqrt{1^{2}-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}.
x=\frac{-1±\sqrt{1-4\times 4\left(-18\right)}}{2\times 4}
Square 1.
x=\frac{-1±\sqrt{1-16\left(-18\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{-1±\sqrt{1+288}}{2\times 4}
Multiply -16 times -18.
x=\frac{-1±\sqrt{289}}{2\times 4}
Add 1 to 288.
x=\frac{-1±17}{2\times 4}
Take the square root of 289.
x=\frac{-1±17}{8}
Multiply 2 times 4.
x=\frac{16}{8}
Now solve the equation x=\frac{-1±17}{8} when ± is plus. Add -1 to 17.
x=2
Divide 16 by 8.
x=-\frac{18}{8}
Now solve the equation x=\frac{-1±17}{8} when ± is minus. Subtract 17 from -1.
x=-\frac{9}{4}
Reduce the fraction \frac{-18}{8} to lowest terms by extracting and canceling out 2.
x=2 x=-\frac{9}{4}
The equation is now solved.
4x^{2}+x=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{4x^{2}+x}{4}=\frac{18}{4}
Divide both sides by 4.
x^{2}+\frac{1}{4}x=\frac{18}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+\frac{1}{4}x=\frac{9}{2}
Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{4}x+\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.
x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{9}{2}+\frac{1}{64}
Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{4}x+\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(x+\frac{1}{8}\right)^{2}=\frac{289}{64}
Factor x^{2}+\frac{1}{4}x+\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(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{289}{64}}
Take the square root of both sides of the equation.
x+\frac{1}{8}=\frac{17}{8} x+\frac{1}{8}=-\frac{17}{8}
Simplify.
x=2 x=-\frac{9}{4}
Subtract \frac{1}{8} from both sides of the equation.