Solve 11b-7=-6b^2

$$11 b - 7 = - 6 b ^ { 2 }$$

Solve for b

$$b = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

$$b=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Add 6b^{2} to both sides.

$$11b-7+6b^{2}=0$$

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

$$6b^{2}+11b-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6b^{2}+ab+bb-7. To find a and b, set up a system to be solved.

$$a+b=11 ab=6\left(-7\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 11.

$$a=-3 b=14$$

Rewrite 6b^{2}+11b-7 as \left(6b^{2}-3b\right)+\left(14b-7\right).

$$\left(6b^{2}-3b\right)+\left(14b-7\right)$$

Factor out 3b in the first and 7 in the second group.

$$3b\left(2b-1\right)+7\left(2b-1\right)$$

Factor out common term 2b-1 by using distributive property.

$$\left(2b-1\right)\left(3b+7\right)$$

To find equation solutions, solve 2b-1=0 and 3b+7=0.

$$b=\frac{1}{2} b=-\frac{7}{3}$$

Steps Using the Quadratic Formula

Add 6b^{2} to both sides.

$$11b-7+6b^{2}=0$$

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.

$$6b^{2}+11b-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 11 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-11±\sqrt{11^{2}-4\times 6\left(-7\right)}}{2\times 6}$$

Square 11.

$$b=\frac{-11±\sqrt{121-4\times 6\left(-7\right)}}{2\times 6}$$

Multiply -4 times 6.

$$b=\frac{-11±\sqrt{121-24\left(-7\right)}}{2\times 6}$$

Multiply -24 times -7.

$$b=\frac{-11±\sqrt{121+168}}{2\times 6}$$

Add 121 to 168.

$$b=\frac{-11±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$b=\frac{-11±17}{2\times 6}$$

Multiply 2 times 6.

$$b=\frac{-11±17}{12}$$

Now solve the equation b=\frac{-11±17}{12} when ± is plus. Add -11 to 17.

$$b=\frac{6}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$b=\frac{1}{2}$$

Now solve the equation b=\frac{-11±17}{12} when ± is minus. Subtract 17 from -11.

$$b=-\frac{28}{12}$$

Reduce the fraction \frac{-28}{12} to lowest terms by extracting and canceling out 4.

$$b=-\frac{7}{3}$$

The equation is now solved.

$$b=\frac{1}{2} b=-\frac{7}{3}$$

Steps for Completing the Square

Add 6b^{2} to both sides.

$$11b-7+6b^{2}=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$11b+6b^{2}=7$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$6b^{2}+11b=7$$

Divide both sides by 6.

$$\frac{6b^{2}+11b}{6}=\frac{7}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$b^{2}+\frac{11}{6}b=\frac{7}{6}$$

Divide \frac{11}{6}, the coefficient of the x term, by 2 to get \frac{11}{12}. Then add the square of \frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}+\frac{11}{6}b+\left(\frac{11}{12}\right)^{2}=\frac{7}{6}+\left(\frac{11}{12}\right)^{2}$$

Square \frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$b^{2}+\frac{11}{6}b+\frac{121}{144}=\frac{7}{6}+\frac{121}{144}$$

Add \frac{7}{6} to \frac{121}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}+\frac{11}{6}b+\frac{121}{144}=\frac{289}{144}$$

Factor b^{2}+\frac{11}{6}b+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b+\frac{11}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+\frac{11}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$b+\frac{11}{12}=\frac{17}{12} b+\frac{11}{12}=-\frac{17}{12}$$

Subtract \frac{11}{12} from both sides of the equation.

$$b=\frac{1}{2} b=-\frac{7}{3}$$