Solve 1/2=4+b^2-6/4b

$$\frac { 1 } { 2 } = \frac { 4 + b ^ { 2 } - 6 } { 4 b }$$

Solve for b

$$b=\sqrt{3}+1\approx 2.732050808$$

$$b=1-\sqrt{3}\approx -0.732050808$$

Steps Using the Quadratic Formula

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4b, the least common multiple of 2,4b.

$$2b=4+b^{2}-6$$

Subtract 6 from 4 to get -2.

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

Subtract -2 from both sides.

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

The opposite of -2 is 2.

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

Subtract b^{2} from both sides.

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

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.

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

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

$$b=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square 2.

$$b=\frac{-2±\sqrt{4-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$b=\frac{-2±\sqrt{4+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$b=\frac{-2±\sqrt{4+8}}{2\left(-1\right)}$$

Add 4 to 8.

$$b=\frac{-2±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$b=\frac{-2±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$b=\frac{-2±2\sqrt{3}}{-2}$$

Now solve the equation b=\frac{-2±2\sqrt{3}}{-2} when ± is plus. Add -2 to 2\sqrt{3}.

$$b=\frac{2\sqrt{3}-2}{-2}$$

Divide -2+2\sqrt{3} by -2.

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

Now solve the equation b=\frac{-2±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -2.

$$b=\frac{-2\sqrt{3}-2}{-2}$$

Divide -2-2\sqrt{3} by -2.

$$b=\sqrt{3}+1$$

The equation is now solved.

$$b=1-\sqrt{3} b=\sqrt{3}+1$$

Steps for Completing the Square

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4b, the least common multiple of 2,4b.

$$2b=4+b^{2}-6$$

Subtract 6 from 4 to get -2.

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

Subtract b^{2} from both sides.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 2 by -1.

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

Divide -2 by -1.

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

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.

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

Add 2 to 1.

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

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}.

$$\left(b-1\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-1\right)^{2}}=\sqrt{3}$$

Simplify.

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

Add 1 to both sides of the equation.

$$b=\sqrt{3}+1 b=1-\sqrt{3}$$