Solve 3w^2=-17w-20

$$3 w ^ { 2 } = - 17 w - 20$$

Solve for w

$$w = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$w=-4$$

Steps Using Factoring By Grouping

Add 17w to both sides.

$$3w^{2}+17w=-20$$

Add 20 to both sides.

$$3w^{2}+17w+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3w^{2}+aw+bw+20. To find a and b, set up a system to be solved.

$$a+b=17 ab=3\times 20=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 17.

$$a=5 b=12$$

Rewrite 3w^{2}+17w+20 as \left(3w^{2}+5w\right)+\left(12w+20\right).

$$\left(3w^{2}+5w\right)+\left(12w+20\right)$$

Factor out w in the first and 4 in the second group.

$$w\left(3w+5\right)+4\left(3w+5\right)$$

Factor out common term 3w+5 by using distributive property.

$$\left(3w+5\right)\left(w+4\right)$$

To find equation solutions, solve 3w+5=0 and w+4=0.

$$w=-\frac{5}{3} w=-4$$

Steps Using the Quadratic Formula

Add 17w to both sides.

$$3w^{2}+17w=-20$$

Add 20 to both sides.

$$3w^{2}+17w+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 17 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-17±\sqrt{17^{2}-4\times 3\times 20}}{2\times 3}$$

Square 17.

$$w=\frac{-17±\sqrt{289-4\times 3\times 20}}{2\times 3}$$

Multiply -4 times 3.

$$w=\frac{-17±\sqrt{289-12\times 20}}{2\times 3}$$

Multiply -12 times 20.

$$w=\frac{-17±\sqrt{289-240}}{2\times 3}$$

Add 289 to -240.

$$w=\frac{-17±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$w=\frac{-17±7}{2\times 3}$$

Multiply 2 times 3.

$$w=\frac{-17±7}{6}$$

Now solve the equation w=\frac{-17±7}{6} when ± is plus. Add -17 to 7.

$$w=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

$$w=-\frac{5}{3}$$

Now solve the equation w=\frac{-17±7}{6} when ± is minus. Subtract 7 from -17.

$$w=-\frac{24}{6}$$

Divide -24 by 6.

$$w=-4$$

The equation is now solved.

$$w=-\frac{5}{3} w=-4$$

Steps for Completing the Square

Add 17w to both sides.

$$3w^{2}+17w=-20$$

Divide both sides by 3.

$$\frac{3w^{2}+17w}{3}=-\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$w^{2}+\frac{17}{3}w=-\frac{20}{3}$$

Divide \frac{17}{3}, the coefficient of the x term, by 2 to get \frac{17}{6}. Then add the square of \frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}+\frac{17}{3}w+\left(\frac{17}{6}\right)^{2}=-\frac{20}{3}+\left(\frac{17}{6}\right)^{2}$$

Square \frac{17}{6} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+\frac{17}{3}w+\frac{289}{36}=-\frac{20}{3}+\frac{289}{36}$$

Add -\frac{20}{3} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$w^{2}+\frac{17}{3}w+\frac{289}{36}=\frac{49}{36}$$

Factor w^{2}+\frac{17}{3}w+\frac{289}{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}.

$$\left(w+\frac{17}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{17}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$w+\frac{17}{6}=\frac{7}{6} w+\frac{17}{6}=-\frac{7}{6}$$

Subtract \frac{17}{6} from both sides of the equation.

$$w=-\frac{5}{3} w=-4$$