Solve for x
x=-7
x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667
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a+b=10 ab=3\left(-77\right)=-231
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-77. To find a and b, set up a system to be solved.
-1,231 -3,77 -7,33 -11,21
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 -231.
-1+231=230 -3+77=74 -7+33=26 -11+21=10
Calculate the sum for each pair.
a=-11 b=21
The solution is the pair that gives sum 10.
\left(3x^{2}-11x\right)+\left(21x-77\right)
Rewrite 3x^{2}+10x-77 as \left(3x^{2}-11x\right)+\left(21x-77\right).
x\left(3x-11\right)+7\left(3x-11\right)
Factor out x in the first and 7 in the second group.
\left(3x-11\right)\left(x+7\right)
Factor out common term 3x-11 by using distributive property.
x=\frac{11}{3} x=-7
To find equation solutions, solve 3x-11=0 and x+7=0.
3x^{2}+10x-77=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{-10±\sqrt{10^{2}-4\times 3\left(-77\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 10 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 3\left(-77\right)}}{2\times 3}
Square 10.
x=\frac{-10±\sqrt{100-12\left(-77\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-10±\sqrt{100+924}}{2\times 3}
Multiply -12 times -77.
x=\frac{-10±\sqrt{1024}}{2\times 3}
Add 100 to 924.
x=\frac{-10±32}{2\times 3}
Take the square root of 1024.
x=\frac{-10±32}{6}
Multiply 2 times 3.
x=\frac{22}{6}
Now solve the equation x=\frac{-10±32}{6} when ± is plus. Add -10 to 32.
x=\frac{11}{3}
Reduce the fraction \frac{22}{6} to lowest terms by extracting and canceling out 2.
x=-\frac{42}{6}
Now solve the equation x=\frac{-10±32}{6} when ± is minus. Subtract 32 from -10.
x=-7
Divide -42 by 6.
x=\frac{11}{3} x=-7
The equation is now solved.
3x^{2}+10x-77=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}+10x-77-\left(-77\right)=-\left(-77\right)
Add 77 to both sides of the equation.
3x^{2}+10x=-\left(-77\right)
Subtracting -77 from itself leaves 0.
3x^{2}+10x=77
Subtract -77 from 0.
\frac{3x^{2}+10x}{3}=\frac{77}{3}
Divide both sides by 3.
x^{2}+\frac{10}{3}x=\frac{77}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{77}{3}+\left(\frac{5}{3}\right)^{2}
Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{77}{3}+\frac{25}{9}
Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{256}{9}
Add \frac{77}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{3}\right)^{2}=\frac{256}{9}
Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. 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{5}{3}\right)^{2}}=\sqrt{\frac{256}{9}}
Take the square root of both sides of the equation.
x+\frac{5}{3}=\frac{16}{3} x+\frac{5}{3}=-\frac{16}{3}
Simplify.
x=\frac{11}{3} x=-7
Subtract \frac{5}{3} from both sides of the equation.
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