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a+b=5 ab=-3\times 2=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
-1,6 -2,3
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 -6.
-1+6=5 -2+3=1
Calculate the sum for each pair.
a=6 b=-1
The solution is the pair that gives sum 5.
\left(-3x^{2}+6x\right)+\left(-x+2\right)
Rewrite -3x^{2}+5x+2 as \left(-3x^{2}+6x\right)+\left(-x+2\right).
3x\left(-x+2\right)-x+2
Factor out 3x in -3x^{2}+6x.
\left(-x+2\right)\left(3x+1\right)
Factor out common term -x+2 by using distributive property.
x=2 x=-\frac{1}{3}
To find equation solutions, solve -x+2=0 and 3x+1=0.
-3x^{2}+5x+2=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\left(-3\right)\times 2}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 5 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\left(-3\right)\times 2}}{2\left(-3\right)}
Square 5.
x=\frac{-5±\sqrt{25+12\times 2}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-5±\sqrt{25+24}}{2\left(-3\right)}
Multiply 12 times 2.
x=\frac{-5±\sqrt{49}}{2\left(-3\right)}
Add 25 to 24.
x=\frac{-5±7}{2\left(-3\right)}
Take the square root of 49.
x=\frac{-5±7}{-6}
Multiply 2 times -3.
x=\frac{2}{-6}
Now solve the equation x=\frac{-5±7}{-6} when ± is plus. Add -5 to 7.
x=-\frac{1}{3}
Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.
x=-\frac{12}{-6}
Now solve the equation x=\frac{-5±7}{-6} when ± is minus. Subtract 7 from -5.
x=2
Divide -12 by -6.
x=-\frac{1}{3} x=2
The equation is now solved.
-3x^{2}+5x+2=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+2-2=-2
Subtract 2 from both sides of the equation.
-3x^{2}+5x=-2
Subtracting 2 from itself leaves 0.
\frac{-3x^{2}+5x}{-3}=-\frac{2}{-3}
Divide both sides by -3.
x^{2}+\frac{5}{-3}x=-\frac{2}{-3}
Dividing by -3 undoes the multiplication by -3.
x^{2}-\frac{5}{3}x=-\frac{2}{-3}
Divide 5 by -3.
x^{2}-\frac{5}{3}x=\frac{2}{3}
Divide -2 by -3.
x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\frac{2}{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{2}{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{49}{36}
Add \frac{2}{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{49}{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{49}{36}}
Take the square root of both sides of the equation.
x-\frac{5}{6}=\frac{7}{6} x-\frac{5}{6}=-\frac{7}{6}
Simplify.
x=2 x=-\frac{1}{3}
Add \frac{5}{6} to both sides of the equation.