Solve for x
x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667
x=1
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a+b=5 ab=3\left(-8\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-8. 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 positive, the positive number has greater absolute value than the negative. 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=-3 b=8
The solution is the pair that gives sum 5.
\left(3x^{2}-3x\right)+\left(8x-8\right)
Rewrite 3x^{2}+5x-8 as \left(3x^{2}-3x\right)+\left(8x-8\right).
3x\left(x-1\right)+8\left(x-1\right)
Factor out 3x in the first and 8 in the second group.
\left(x-1\right)\left(3x+8\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{8}{3}
To find equation solutions, solve x-1=0 and 3x+8=0.
3x^{2}+5x-8=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{-5±\sqrt{5^{2}-4\times 3\left(-8\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 3\left(-8\right)}}{2\times 3}
Square 5.
x=\frac{-5±\sqrt{25-12\left(-8\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-5±\sqrt{25+96}}{2\times 3}
Multiply -12 times -8.
x=\frac{-5±\sqrt{121}}{2\times 3}
Add 25 to 96.
x=\frac{-5±11}{2\times 3}
Take the square root of 121.
x=\frac{-5±11}{6}
Multiply 2 times 3.
x=\frac{6}{6}
Now solve the equation x=\frac{-5±11}{6} when ± is plus. Add -5 to 11.
x=1
Divide 6 by 6.
x=-\frac{16}{6}
Now solve the equation x=\frac{-5±11}{6} when ± is minus. Subtract 11 from -5.
x=-\frac{8}{3}
Reduce the fraction \frac{-16}{6} to lowest terms by extracting and canceling out 2.
x=1 x=-\frac{8}{3}
The equation is now solved.
3x^{2}+5x-8=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}+5x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
3x^{2}+5x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
3x^{2}+5x=8
Subtract -8 from 0.
\frac{3x^{2}+5x}{3}=\frac{8}{3}
Divide both sides by 3.
x^{2}+\frac{5}{3}x=\frac{8}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=\frac{8}{3}+\left(\frac{5}{6}\right)^{2}
Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{8}{3}+\frac{25}{36}
Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{121}{36}
Add \frac{8}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{6}\right)^{2}=\frac{121}{36}
Factor x^{2}+\frac{5}{3}x+\frac{25}{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{5}{6}\right)^{2}}=\sqrt{\frac{121}{36}}
Take the square root of both sides of the equation.
x+\frac{5}{6}=\frac{11}{6} x+\frac{5}{6}=-\frac{11}{6}
Simplify.
x=1 x=-\frac{8}{3}
Subtract \frac{5}{6} from both sides of the equation.
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Linear equation
y = 3x + 4
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Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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