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

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

Solve for b (complex solution)

$$b=\sqrt{11}-1\approx 2.31662479$$

$$b=-\left(\sqrt{11}+1\right)\approx -4.31662479$$

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.

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

Add 4 and 6 to get 10.

$$2b=10-b^{2}$$

Subtract 10 from both sides.

$$2b-10=-b^{2}$$

Add b^{2} to both sides.

$$2b-10+b^{2}=0$$

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.

$$b^{2}+2b-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-2±\sqrt{2^{2}-4\left(-10\right)}}{2}$$

Square 2.

$$b=\frac{-2±\sqrt{4-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$b=\frac{-2±\sqrt{4+40}}{2}$$

Add 4 to 40.

$$b=\frac{-2±\sqrt{44}}{2}$$

Take the square root of 44.

$$b=\frac{-2±2\sqrt{11}}{2}$$

Now solve the equation b=\frac{-2±2\sqrt{11}}{2} when ± is plus. Add -2 to 2\sqrt{11}.

$$b=\frac{2\sqrt{11}-2}{2}$$

Divide -2+2\sqrt{11} by 2.

$$b=\sqrt{11}-1$$

Now solve the equation b=\frac{-2±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -2.

$$b=\frac{-2\sqrt{11}-2}{2}$$

Divide -2-2\sqrt{11} by 2.

$$b=-\sqrt{11}-1$$

The equation is now solved.

$$b=\sqrt{11}-1 b=-\sqrt{11}-1$$

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.

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

Add 4 and 6 to get 10.

$$2b=10-b^{2}$$

Add b^{2} to both sides.

$$2b+b^{2}=10$$

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.

$$b^{2}+2b=10$$

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.

$$b^{2}+2b+1^{2}=10+1^{2}$$

Square 1.

$$b^{2}+2b+1=10+1$$

Add 10 to 1.

$$b^{2}+2b+1=11$$

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}.

$$\left(b+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$b+1=\sqrt{11} b+1=-\sqrt{11}$$

Subtract 1 from both sides of the equation.

$$b=\sqrt{11}-1 b=-\sqrt{11}-1$$

Solve for b

$$b=\sqrt{11}-1\approx 2.31662479$$

$$b=-\sqrt{11}-1\approx -4.31662479$$

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.

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

Add 4 and 6 to get 10.

$$2b=10-b^{2}$$

Subtract 10 from both sides.

$$2b-10=-b^{2}$$

Add b^{2} to both sides.

$$2b-10+b^{2}=0$$

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.

$$b^{2}+2b-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-2±\sqrt{2^{2}-4\left(-10\right)}}{2}$$

Square 2.

$$b=\frac{-2±\sqrt{4-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$b=\frac{-2±\sqrt{4+40}}{2}$$

Add 4 to 40.

$$b=\frac{-2±\sqrt{44}}{2}$$

Take the square root of 44.

$$b=\frac{-2±2\sqrt{11}}{2}$$

Now solve the equation b=\frac{-2±2\sqrt{11}}{2} when ± is plus. Add -2 to 2\sqrt{11}.

$$b=\frac{2\sqrt{11}-2}{2}$$

Divide -2+2\sqrt{11} by 2.

$$b=\sqrt{11}-1$$

Now solve the equation b=\frac{-2±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from -2.

$$b=\frac{-2\sqrt{11}-2}{2}$$

Divide -2-2\sqrt{11} by 2.

$$b=-\sqrt{11}-1$$

The equation is now solved.

$$b=\sqrt{11}-1 b=-\sqrt{11}-1$$

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.

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

Add 4 and 6 to get 10.

$$2b=10-b^{2}$$

Add b^{2} to both sides.

$$2b+b^{2}=10$$

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.

$$b^{2}+2b=10$$

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.

$$b^{2}+2b+1^{2}=10+1^{2}$$

Square 1.

$$b^{2}+2b+1=10+1$$

Add 10 to 1.

$$b^{2}+2b+1=11$$

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}.

$$\left(b+1\right)^{2}=11$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b+1\right)^{2}}=\sqrt{11}$$

Simplify.

$$b+1=\sqrt{11} b+1=-\sqrt{11}$$

Subtract 1 from both sides of the equation.

$$b=\sqrt{11}-1 b=-\sqrt{11}-1$$