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3w^{2}+5w+2=0
Add 2 to both sides.
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 3w^{2}+aw+bw+2. To find a and b, set up a system to be solved.
1,6 2,3
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.
1+6=7 2+3=5
Calculate the sum for each pair.
a=2 b=3
The solution is the pair that gives sum 5.
\left(3w^{2}+2w\right)+\left(3w+2\right)
Rewrite 3w^{2}+5w+2 as \left(3w^{2}+2w\right)+\left(3w+2\right).
w\left(3w+2\right)+3w+2
Factor out w in 3w^{2}+2w.
\left(3w+2\right)\left(w+1\right)
Factor out common term 3w+2 by using distributive property.
w=-\frac{2}{3} w=-1
To find equation solutions, solve 3w+2=0 and w+1=0.
3w^{2}+5w=-2
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.
3w^{2}+5w-\left(-2\right)=-2-\left(-2\right)
Add 2 to both sides of the equation.
3w^{2}+5w-\left(-2\right)=0
Subtracting -2 from itself leaves 0.
3w^{2}+5w+2=0
Subtract -2 from 0.
w=\frac{-5±\sqrt{5^{2}-4\times 3\times 2}}{2\times 3}
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}.
w=\frac{-5±\sqrt{25-4\times 3\times 2}}{2\times 3}
Square 5.
w=\frac{-5±\sqrt{25-12\times 2}}{2\times 3}
Multiply -4 times 3.
w=\frac{-5±\sqrt{25-24}}{2\times 3}
Multiply -12 times 2.
w=\frac{-5±\sqrt{1}}{2\times 3}
Add 25 to -24.
w=\frac{-5±1}{2\times 3}
Take the square root of 1.
w=\frac{-5±1}{6}
Multiply 2 times 3.
w=-\frac{4}{6}
Now solve the equation w=\frac{-5±1}{6} when ± is plus. Add -5 to 1.
w=-\frac{2}{3}
Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.
w=-\frac{6}{6}
Now solve the equation w=\frac{-5±1}{6} when ± is minus. Subtract 1 from -5.
w=-1
Divide -6 by 6.
w=-\frac{2}{3} w=-1
The equation is now solved.
3w^{2}+5w=-2
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{3w^{2}+5w}{3}=-\frac{2}{3}
Divide both sides by 3.
w^{2}+\frac{5}{3}w=-\frac{2}{3}
Dividing by 3 undoes the multiplication by 3.
w^{2}+\frac{5}{3}w+\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.
w^{2}+\frac{5}{3}w+\frac{25}{36}=-\frac{2}{3}+\frac{25}{36}
Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.
w^{2}+\frac{5}{3}w+\frac{25}{36}=\frac{1}{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(w+\frac{5}{6}\right)^{2}=\frac{1}{36}
Factor w^{2}+\frac{5}{3}w+\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(w+\frac{5}{6}\right)^{2}}=\sqrt{\frac{1}{36}}
Take the square root of both sides of the equation.
w+\frac{5}{6}=\frac{1}{6} w+\frac{5}{6}=-\frac{1}{6}
Simplify.
w=-\frac{2}{3} w=-1
Subtract \frac{5}{6} from both sides of the equation.