Solve b^2+5b-35=3b

$$b ^ { 2 } + 5 b - 35 = 3 b$$

Solve for b

$$b=-7$$

$$b=5$$

Steps Using Factoring

Subtract 3b from both sides.

$$b^{2}+5b-35-3b=0$$

Combine 5b and -3b to get 2b.

$$b^{2}+2b-35=0$$

To solve the equation, factor b^{2}+2b-35 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

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

$$\left(b-5\right)\left(b+7\right)$$

To find equation solutions, solve b-5=0 and b+7=0.

$$b=5 b=-7$$

Steps Using Factoring By Grouping

Subtract 3b from both sides.

$$b^{2}+5b-35-3b=0$$

Combine 5b and -3b to get 2b.

$$b^{2}+2b-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb-35. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-35\right)=-35$$

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 -35.

$$-1,35 -5,7$$

Calculate the sum for each pair.

$$-1+35=34 -5+7=2$$

The solution is the pair that gives sum 2.

$$a=-5 b=7$$

Rewrite b^{2}+2b-35 as \left(b^{2}-5b\right)+\left(7b-35\right).

$$\left(b^{2}-5b\right)+\left(7b-35\right)$$

Factor out b in the first and 7 in the second group.

$$b\left(b-5\right)+7\left(b-5\right)$$

Factor out common term b-5 by using distributive property.

$$\left(b-5\right)\left(b+7\right)$$

To find equation solutions, solve b-5=0 and b+7=0.

$$b=5 b=-7$$

Steps Using the Quadratic Formula

Subtract 3b from both sides.

$$b^{2}+5b-35-3b=0$$

Combine 5b and -3b to get 2b.

$$b^{2}+2b-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-2±\sqrt{2^{2}-4\left(-35\right)}}{2}$$

Square 2.

$$b=\frac{-2±\sqrt{4-4\left(-35\right)}}{2}$$

Multiply -4 times -35.

$$b=\frac{-2±\sqrt{4+140}}{2}$$

Add 4 to 140.

$$b=\frac{-2±\sqrt{144}}{2}$$

Take the square root of 144.

$$b=\frac{-2±12}{2}$$

Now solve the equation b=\frac{-2±12}{2} when ± is plus. Add -2 to 12.

$$b=\frac{10}{2}$$

Divide 10 by 2.

$$b=5$$

Now solve the equation b=\frac{-2±12}{2} when ± is minus. Subtract 12 from -2.

$$b=-\frac{14}{2}$$

Divide -14 by 2.

$$b=-7$$

The equation is now solved.

$$b=5 b=-7$$

Steps for Completing the Square

Subtract 3b from both sides.

$$b^{2}+5b-35-3b=0$$

Combine 5b and -3b to get 2b.

$$b^{2}+2b-35=0$$

Add 35 to both sides. Anything plus zero gives itself.

$$b^{2}+2b=35$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+2b+1^{2}=35+1^{2}$$

Square 1.

$$b^{2}+2b+1=35+1$$

Add 35 to 1.

$$b^{2}+2b+1=36$$

Factor b^{2}+2b+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+1\right)^{2}}=\sqrt{36}$$

Simplify.

$$b+1=6 b+1=-6$$

Subtract 1 from both sides of the equation.

$$b=5 b=-7$$