Solve 3w^2+6=-11w

$$3 w ^ { 2 } + 6 = - 11 w$$

Solve for w

$$w=-3$$

$$w=-\frac{2}{3}\approx -0.666666667$$

Steps Using Factoring By Grouping

Add 11w to both sides.

$$3w^{2}+6+11w=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3w^{2}+11w+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3w^{2}+aw+bw+6. To find a and b, set up a system to be solved.

$$a+b=11 ab=3\times 6=18$$

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 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 11.

$$a=2 b=9$$

Rewrite 3w^{2}+11w+6 as \left(3w^{2}+2w\right)+\left(9w+6\right).

$$\left(3w^{2}+2w\right)+\left(9w+6\right)$$

Factor out w in the first and 3 in the second group.

$$w\left(3w+2\right)+3\left(3w+2\right)$$

Factor out common term 3w+2 by using distributive property.

$$\left(3w+2\right)\left(w+3\right)$$

To find equation solutions, solve 3w+2=0 and w+3=0.

$$w=-\frac{2}{3} w=-3$$

Steps Using the Quadratic Formula

Add 11w to both sides.

$$3w^{2}+6+11w=0$$

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.

$$3w^{2}+11w+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 11 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-11±\sqrt{11^{2}-4\times 3\times 6}}{2\times 3}$$

Square 11.

$$w=\frac{-11±\sqrt{121-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$w=\frac{-11±\sqrt{121-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$w=\frac{-11±\sqrt{121-72}}{2\times 3}$$

Add 121 to -72.

$$w=\frac{-11±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$w=\frac{-11±7}{2\times 3}$$

Multiply 2 times 3.

$$w=\frac{-11±7}{6}$$

Now solve the equation w=\frac{-11±7}{6} when ± is plus. Add -11 to 7.

$$w=-\frac{4}{6}$$

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$w=-\frac{2}{3}$$

Now solve the equation w=\frac{-11±7}{6} when ± is minus. Subtract 7 from -11.

$$w=-\frac{18}{6}$$

Divide -18 by 6.

$$w=-3$$

The equation is now solved.

$$w=-\frac{2}{3} w=-3$$

Steps for Completing the Square

Add 11w to both sides.

$$3w^{2}+6+11w=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$3w^{2}+11w=-6$$

Divide both sides by 3.

$$\frac{3w^{2}+11w}{3}=-\frac{6}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$w^{2}+\frac{11}{3}w=-\frac{6}{3}$$

Divide -6 by 3.

$$w^{2}+\frac{11}{3}w=-2$$

Divide \frac{11}{3}, the coefficient of the x term, by 2 to get \frac{11}{6}. Then add the square of \frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}+\frac{11}{3}w+\left(\frac{11}{6}\right)^{2}=-2+\left(\frac{11}{6}\right)^{2}$$

Square \frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+\frac{11}{3}w+\frac{121}{36}=-2+\frac{121}{36}$$

Add -2 to \frac{121}{36}.

$$w^{2}+\frac{11}{3}w+\frac{121}{36}=\frac{49}{36}$$

Factor w^{2}+\frac{11}{3}w+\frac{121}{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}.

$$\left(w+\frac{11}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{11}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$w+\frac{11}{6}=\frac{7}{6} w+\frac{11}{6}=-\frac{7}{6}$$

Subtract \frac{11}{6} from both sides of the equation.

$$w=-\frac{2}{3} w=-3$$