Solve 2b^2-105=-11b

$$2 b ^ { 2 } - 105 = - 11 b$$

Solve for b

$$b = -\frac{21}{2} = -10\frac{1}{2} = -10.5$$

$$b=5$$

Steps Using Factoring By Grouping

Add 11b to both sides.

$$2b^{2}-105+11b=0$$

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

$$2b^{2}+11b-105=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2b^{2}+ab+bb-105. To find a and b, set up a system to be solved.

$$a+b=11 ab=2\left(-105\right)=-210$$

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 -210.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Calculate the sum for each pair.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 11.

$$a=-10 b=21$$

Rewrite 2b^{2}+11b-105 as \left(2b^{2}-10b\right)+\left(21b-105\right).

$$\left(2b^{2}-10b\right)+\left(21b-105\right)$$

Factor out 2b in the first and 21 in the second group.

$$2b\left(b-5\right)+21\left(b-5\right)$$

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

$$\left(b-5\right)\left(2b+21\right)$$

To find equation solutions, solve b-5=0 and 2b+21=0.

$$b=5 b=-\frac{21}{2}$$

Steps Using the Quadratic Formula

Add 11b to both sides.

$$2b^{2}-105+11b=0$$

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.

$$2b^{2}+11b-105=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 11 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-11±\sqrt{11^{2}-4\times 2\left(-105\right)}}{2\times 2}$$

Square 11.

$$b=\frac{-11±\sqrt{121-4\times 2\left(-105\right)}}{2\times 2}$$

Multiply -4 times 2.

$$b=\frac{-11±\sqrt{121-8\left(-105\right)}}{2\times 2}$$

Multiply -8 times -105.

$$b=\frac{-11±\sqrt{121+840}}{2\times 2}$$

Add 121 to 840.

$$b=\frac{-11±\sqrt{961}}{2\times 2}$$

Take the square root of 961.

$$b=\frac{-11±31}{2\times 2}$$

Multiply 2 times 2.

$$b=\frac{-11±31}{4}$$

Now solve the equation b=\frac{-11±31}{4} when ± is plus. Add -11 to 31.

$$b=\frac{20}{4}$$

Divide 20 by 4.

$$b=5$$

Now solve the equation b=\frac{-11±31}{4} when ± is minus. Subtract 31 from -11.

$$b=-\frac{42}{4}$$

Reduce the fraction \frac{-42}{4} to lowest terms by extracting and canceling out 2.

$$b=-\frac{21}{2}$$

The equation is now solved.

$$b=5 b=-\frac{21}{2}$$

Steps for Completing the Square

Add 11b to both sides.

$$2b^{2}-105+11b=0$$

Add 105 to both sides. Anything plus zero gives itself.

$$2b^{2}+11b=105$$

Divide both sides by 2.

$$\frac{2b^{2}+11b}{2}=\frac{105}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$b^{2}+\frac{11}{2}b=\frac{105}{2}$$

Divide \frac{11}{2}, the coefficient of the x term, by 2 to get \frac{11}{4}. Then add the square of \frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+\frac{11}{2}b+\left(\frac{11}{4}\right)^{2}=\frac{105}{2}+\left(\frac{11}{4}\right)^{2}$$

Square \frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$b^{2}+\frac{11}{2}b+\frac{121}{16}=\frac{105}{2}+\frac{121}{16}$$

Add \frac{105}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}+\frac{11}{2}b+\frac{121}{16}=\frac{961}{16}$$

Factor b^{2}+\frac{11}{2}b+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+\frac{11}{4}\right)^{2}=\frac{961}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{11}{4}\right)^{2}}=\sqrt{\frac{961}{16}}$$

Simplify.

$$b+\frac{11}{4}=\frac{31}{4} b+\frac{11}{4}=-\frac{31}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$b=5 b=-\frac{21}{2}$$