Solve b+1=-1/2(b+1)^2

$$b + 1 = - \frac { 1 } { 2 } ( b + 1 ) ^ { 2 }$$

Solve for b

$$b=-3$$

$$b=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(b+1\right)^{2}.

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

Use the distributive property to multiply -\frac{1}{2} by b^{2}+2b+1.

$$b+1=-\frac{1}{2}b^{2}-b-\frac{1}{2}$$

Add \frac{1}{2}b^{2} to both sides.

$$b+1+\frac{1}{2}b^{2}=-b-\frac{1}{2}$$

Add b to both sides.

$$b+1+\frac{1}{2}b^{2}+b=-\frac{1}{2}$$

Combine b and b to get 2b.

$$2b+1+\frac{1}{2}b^{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides.

$$2b+1+\frac{1}{2}b^{2}+\frac{1}{2}=0$$

Add 1 and \frac{1}{2} to get \frac{3}{2}.

$$2b+\frac{3}{2}+\frac{1}{2}b^{2}=0$$

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.

$$\frac{1}{2}b^{2}+2b+\frac{3}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, 2 for b, and \frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{2}\times \frac{3}{2}}}{2\times \frac{1}{2}}$$

Square 2.

$$b=\frac{-2±\sqrt{4-4\times \frac{1}{2}\times \frac{3}{2}}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$b=\frac{-2±\sqrt{4-2\times \frac{3}{2}}}{2\times \frac{1}{2}}$$

Multiply -2 times \frac{3}{2}.

$$b=\frac{-2±\sqrt{4-3}}{2\times \frac{1}{2}}$$

Add 4 to -3.

$$b=\frac{-2±\sqrt{1}}{2\times \frac{1}{2}}$$

Take the square root of 1.

$$b=\frac{-2±1}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$b=\frac{-2±1}{1}$$

Now solve the equation b=\frac{-2±1}{1} when ± is plus. Add -2 to 1.

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

Divide -1 by 1.

$$b=-1$$

Now solve the equation b=\frac{-2±1}{1} when ± is minus. Subtract 1 from -2.

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

Divide -3 by 1.

$$b=-3$$

The equation is now solved.

$$b=-1 b=-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(b+1\right)^{2}.

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

Use the distributive property to multiply -\frac{1}{2} by b^{2}+2b+1.

$$b+1=-\frac{1}{2}b^{2}-b-\frac{1}{2}$$

Add \frac{1}{2}b^{2} to both sides.

$$b+1+\frac{1}{2}b^{2}=-b-\frac{1}{2}$$

Add b to both sides.

$$b+1+\frac{1}{2}b^{2}+b=-\frac{1}{2}$$

Combine b and b to get 2b.

$$2b+1+\frac{1}{2}b^{2}=-\frac{1}{2}$$

Subtract 1 from both sides.

$$2b+\frac{1}{2}b^{2}=-\frac{1}{2}-1$$

Subtract 1 from -\frac{1}{2} to get -\frac{3}{2}.

$$2b+\frac{1}{2}b^{2}=-\frac{3}{2}$$

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.

$$\frac{1}{2}b^{2}+2b=-\frac{3}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}b^{2}+2b}{\frac{1}{2}}=-\frac{\frac{3}{2}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$b^{2}+\frac{2}{\frac{1}{2}}b=-\frac{\frac{3}{2}}{\frac{1}{2}}$$

Divide 2 by \frac{1}{2} by multiplying 2 by the reciprocal of \frac{1}{2}.

$$b^{2}+4b=-\frac{\frac{3}{2}}{\frac{1}{2}}$$

Divide -\frac{3}{2} by \frac{1}{2} by multiplying -\frac{3}{2} by the reciprocal of \frac{1}{2}.

$$b^{2}+4b=-3$$

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

$$b^{2}+4b+2^{2}=-3+2^{2}$$

Square 2.

$$b^{2}+4b+4=-3+4$$

Add -3 to 4.

$$b^{2}+4b+4=1$$

Factor b^{2}+4b+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$b+2=1 b+2=-1$$

Subtract 2 from both sides of the equation.

$$b=-1 b=-3$$