Solve b^2=15+2b

$$b ^ { 2 } = 15 + 2 b$$

Solve for b

$$b=-3$$

$$b=5$$

Steps Using Factoring

Subtract 15 from both sides.

$$b^{2}-15=2b$$

Subtract 2b from both sides.

$$b^{2}-15-2b=0$$

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

$$b^{2}-2b-15=0$$

To solve the equation, factor b^{2}-2b-15 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.

$$a+b=-2 ab=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

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

$$\left(b-5\right)\left(b+3\right)$$

To find equation solutions, solve b-5=0 and b+3=0.

$$b=5 b=-3$$

Steps Using Factoring By Grouping

Subtract 15 from both sides.

$$b^{2}-15=2b$$

Subtract 2b from both sides.

$$b^{2}-15-2b=0$$

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

$$b^{2}-2b-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb-15. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-15\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

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

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

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

$$b\left(b-5\right)+3\left(b-5\right)$$

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

$$\left(b-5\right)\left(b+3\right)$$

To find equation solutions, solve b-5=0 and b+3=0.

$$b=5 b=-3$$

Steps Using the Quadratic Formula

Subtract 15 from both sides.

$$b^{2}-15=2b$$

Subtract 2b from both sides.

$$b^{2}-15-2b=0$$

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.

$$b^{2}-2b-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-15\right)}}{2}$$

Square -2.

$$b=\frac{-\left(-2\right)±\sqrt{4-4\left(-15\right)}}{2}$$

Multiply -4 times -15.

$$b=\frac{-\left(-2\right)±\sqrt{4+60}}{2}$$

Add 4 to 60.

$$b=\frac{-\left(-2\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$b=\frac{-\left(-2\right)±8}{2}$$

The opposite of -2 is 2.

$$b=\frac{2±8}{2}$$

Now solve the equation b=\frac{2±8}{2} when ± is plus. Add 2 to 8.

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

Divide 10 by 2.

$$b=5$$

Now solve the equation b=\frac{2±8}{2} when ± is minus. Subtract 8 from 2.

$$b=-\frac{6}{2}$$

Divide -6 by 2.

$$b=-3$$

The equation is now solved.

$$b=5 b=-3$$

Steps for Completing the Square

Subtract 2b from both sides.

$$b^{2}-2b=15$$

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.

$$b^{2}-2b+1=15+1$$

Add 15 to 1.

$$b^{2}-2b+1=16$$

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}.

$$\left(b-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$b-1=4 b-1=-4$$

Add 1 to both sides of the equation.

$$b=5 b=-3$$