Solve 36=w^2+5w

$$36 = w ^ { 2 } + 5 w$$

Solve for w

$$w=-9$$

$$w=4$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$w^{2}+5w=36$$

Subtract 36 from both sides.

$$w^{2}+5w-36=0$$

To solve the equation, factor w^{2}+5w-36 using formula w^{2}+\left(a+b\right)w+ab=\left(w+a\right)\left(w+b\right). To find a and b, set up a system to be solved.

$$a+b=5 ab=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite factored expression \left(w+a\right)\left(w+b\right) using the obtained values.

$$\left(w-4\right)\left(w+9\right)$$

To find equation solutions, solve w-4=0 and w+9=0.

$$w=4 w=-9$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$w^{2}+5w=36$$

Subtract 36 from both sides.

$$w^{2}+5w-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw-36. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-36\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite w^{2}+5w-36 as \left(w^{2}-4w\right)+\left(9w-36\right).

$$\left(w^{2}-4w\right)+\left(9w-36\right)$$

Factor out w in the first and 9 in the second group.

$$w\left(w-4\right)+9\left(w-4\right)$$

Factor out common term w-4 by using distributive property.

$$\left(w-4\right)\left(w+9\right)$$

To find equation solutions, solve w-4=0 and w+9=0.

$$w=4 w=-9$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$w^{2}+5w=36$$

Subtract 36 from both sides.

$$w^{2}+5w-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-5±\sqrt{5^{2}-4\left(-36\right)}}{2}$$

Square 5.

$$w=\frac{-5±\sqrt{25-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$w=\frac{-5±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$w=\frac{-5±\sqrt{169}}{2}$$

Take the square root of 169.

$$w=\frac{-5±13}{2}$$

Now solve the equation w=\frac{-5±13}{2} when ± is plus. Add -5 to 13.

$$w=\frac{8}{2}$$

Divide 8 by 2.

$$w=4$$

Now solve the equation w=\frac{-5±13}{2} when ± is minus. Subtract 13 from -5.

$$w=-\frac{18}{2}$$

Divide -18 by 2.

$$w=-9$$

The equation is now solved.

$$w=4 w=-9$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$w^{2}+5w=36$$

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

$$w^{2}+5w+\left(\frac{5}{2}\right)^{2}=36+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$w^{2}+5w+\frac{25}{4}=36+\frac{25}{4}$$

Add 36 to \frac{25}{4}.

$$w^{2}+5w+\frac{25}{4}=\frac{169}{4}$$

Factor w^{2}+5w+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(w+\frac{5}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w+\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$w+\frac{5}{2}=\frac{13}{2} w+\frac{5}{2}=-\frac{13}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$w=4 w=-9$$