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a+b=-1 ab=3\left(-234\right)=-702
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-234. To find a and b, set up a system to be solved.
1,-702 2,-351 3,-234 6,-117 9,-78 13,-54 18,-39 26,-27
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 -702.
1-702=-701 2-351=-349 3-234=-231 6-117=-111 9-78=-69 13-54=-41 18-39=-21 26-27=-1
Calculate the sum for each pair.
a=-27 b=26
The solution is the pair that gives sum -1.
\left(3x^{2}-27x\right)+\left(26x-234\right)
Rewrite 3x^{2}-x-234 as \left(3x^{2}-27x\right)+\left(26x-234\right).
3x\left(x-9\right)+26\left(x-9\right)
Factor out 3x in the first and 26 in the second group.
\left(x-9\right)\left(3x+26\right)
Factor out common term x-9 by using distributive property.
x=9 x=-\frac{26}{3}
To find equation solutions, solve x-9=0 and 3x+26=0.
3x^{2}-x-234=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.
x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-234\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -234 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-12\left(-234\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-1\right)±\sqrt{1+2808}}{2\times 3}
Multiply -12 times -234.
x=\frac{-\left(-1\right)±\sqrt{2809}}{2\times 3}
Add 1 to 2808.
x=\frac{-\left(-1\right)±53}{2\times 3}
Take the square root of 2809.
x=\frac{1±53}{2\times 3}
The opposite of -1 is 1.
x=\frac{1±53}{6}
Multiply 2 times 3.
x=\frac{54}{6}
Now solve the equation x=\frac{1±53}{6} when ± is plus. Add 1 to 53.
x=9
Divide 54 by 6.
x=-\frac{52}{6}
Now solve the equation x=\frac{1±53}{6} when ± is minus. Subtract 53 from 1.
x=-\frac{26}{3}
Reduce the fraction \frac{-52}{6} to lowest terms by extracting and canceling out 2.
x=9 x=-\frac{26}{3}
The equation is now solved.
3x^{2}-x-234=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.
3x^{2}-x-234-\left(-234\right)=-\left(-234\right)
Add 234 to both sides of the equation.
3x^{2}-x=-\left(-234\right)
Subtracting -234 from itself leaves 0.
3x^{2}-x=234
Subtract -234 from 0.
\frac{3x^{2}-x}{3}=\frac{234}{3}
Divide both sides by 3.
x^{2}-\frac{1}{3}x=\frac{234}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{1}{3}x=78
Divide 234 by 3.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=78+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{3}x+\frac{1}{36}=78+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{2809}{36}
Add 78 to \frac{1}{36}.
\left(x-\frac{1}{6}\right)^{2}=\frac{2809}{36}
Factor x^{2}-\frac{1}{3}x+\frac{1}{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(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{2809}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{53}{6} x-\frac{1}{6}=-\frac{53}{6}
Simplify.
x=9 x=-\frac{26}{3}
Add \frac{1}{6} to both sides of the equation.