Solve 5w^2=-16w-3

$$5 w ^ { 2 } = - 16 w - 3$$

Solve for w

$$w=-3$$

$$w=-\frac{1}{5}=-0.2$$

Steps Using Factoring By Grouping

Add 16w to both sides.

$$5w^{2}+16w=-3$$

Add 3 to both sides.

$$5w^{2}+16w+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5w^{2}+aw+bw+3. To find a and b, set up a system to be solved.

$$a+b=16 ab=5\times 3=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 16.

$$a=1 b=15$$

Rewrite 5w^{2}+16w+3 as \left(5w^{2}+w\right)+\left(15w+3\right).

$$\left(5w^{2}+w\right)+\left(15w+3\right)$$

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

$$w\left(5w+1\right)+3\left(5w+1\right)$$

Factor out common term 5w+1 by using distributive property.

$$\left(5w+1\right)\left(w+3\right)$$

To find equation solutions, solve 5w+1=0 and w+3=0.

$$w=-\frac{1}{5} w=-3$$

Steps Using the Quadratic Formula

Add 16w to both sides.

$$5w^{2}+16w=-3$$

Add 3 to both sides.

$$5w^{2}+16w+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 16 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-16±\sqrt{16^{2}-4\times 5\times 3}}{2\times 5}$$

Square 16.

$$w=\frac{-16±\sqrt{256-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 5.

$$w=\frac{-16±\sqrt{256-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$w=\frac{-16±\sqrt{256-60}}{2\times 5}$$

Add 256 to -60.

$$w=\frac{-16±\sqrt{196}}{2\times 5}$$

Take the square root of 196.

$$w=\frac{-16±14}{2\times 5}$$

Multiply 2 times 5.

$$w=\frac{-16±14}{10}$$

Now solve the equation w=\frac{-16±14}{10} when ± is plus. Add -16 to 14.

$$w=-\frac{2}{10}$$

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

$$w=-\frac{1}{5}$$

Now solve the equation w=\frac{-16±14}{10} when ± is minus. Subtract 14 from -16.

$$w=-\frac{30}{10}$$

Divide -30 by 10.

$$w=-3$$

The equation is now solved.

$$w=-\frac{1}{5} w=-3$$

Steps for Completing the Square

Add 16w to both sides.

$$5w^{2}+16w=-3$$

Divide both sides by 5.

$$\frac{5w^{2}+16w}{5}=-\frac{3}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$w^{2}+\frac{16}{5}w=-\frac{3}{5}$$

Divide \frac{16}{5}, the coefficient of the x term, by 2 to get \frac{8}{5}. Then add the square of \frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}+\frac{16}{5}w+\left(\frac{8}{5}\right)^{2}=-\frac{3}{5}+\left(\frac{8}{5}\right)^{2}$$

Square \frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+\frac{16}{5}w+\frac{64}{25}=-\frac{3}{5}+\frac{64}{25}$$

Add -\frac{3}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$w^{2}+\frac{16}{5}w+\frac{64}{25}=\frac{49}{25}$$

Factor w^{2}+\frac{16}{5}w+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(w+\frac{8}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{8}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$w+\frac{8}{5}=\frac{7}{5} w+\frac{8}{5}=-\frac{7}{5}$$

Subtract \frac{8}{5} from both sides of the equation.

$$w=-\frac{1}{5} w=-3$$