Solve 0.38^2+B+(B/2)^2=0.38

$$0.38 ^ { 2 } + B + ( \frac { B } { 2 } ) ^ { 2 } = 0.38$$

Solve for B

$$B=\frac{\sqrt{3089}}{25}-2\approx 0.223150917$$

$$B=-\frac{\sqrt{3089}}{25}-2\approx -4.223150917$$

Steps Using the Quadratic Formula

Calculate 0.38 to the power of 2 and get 0.1444.

$$0.1444+B+\left(\frac{B}{2}\right)^{2}=0.38$$

To raise \frac{B}{2} to a power, raise both numerator and denominator to the power and then divide.

$$0.1444+B+\frac{B^{2}}{2^{2}}=0.38$$

To add or subtract expressions, expand them to make their denominators the same. Multiply B times \frac{2^{2}}{2^{2}}.

$$0.1444+\frac{B\times 2^{2}}{2^{2}}+\frac{B^{2}}{2^{2}}=0.38$$

Since \frac{B\times 2^{2}}{2^{2}} and \frac{B^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$0.1444+\frac{B\times 2^{2}+B^{2}}{2^{2}}=0.38$$

Do the multiplications in B\times 2^{2}+B^{2}.

$$0.1444+\frac{4B+B^{2}}{2^{2}}=0.38$$

Calculate 2 to the power of 2 and get 4.

$$0.1444+\frac{4B+B^{2}}{4}=0.38$$

Divide each term of 4B+B^{2} by 4 to get B+\frac{1}{4}B^{2}.

$$0.1444+B+\frac{1}{4}B^{2}=0.38$$

Subtract 0.38 from both sides.

$$0.1444+B+\frac{1}{4}B^{2}-0.38=0$$

Subtract 0.38 from 0.1444 to get -0.2356.

$$-0.2356+B+\frac{1}{4}B^{2}=0$$

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.

$$\frac{1}{4}B^{2}+B-0.2356=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, 1 for b, and -0.2356 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$B=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{4}\left(-0.2356\right)}}{2\times \frac{1}{4}}$$

Square 1.

$$B=\frac{-1±\sqrt{1-4\times \frac{1}{4}\left(-0.2356\right)}}{2\times \frac{1}{4}}$$

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

$$B=\frac{-1±\sqrt{1-\left(-0.2356\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -0.2356.

$$B=\frac{-1±\sqrt{1+0.2356}}{2\times \frac{1}{4}}$$

Add 1 to 0.2356.

$$B=\frac{-1±\sqrt{1.2356}}{2\times \frac{1}{4}}$$

Take the square root of 1.2356.

$$B=\frac{-1±\frac{\sqrt{3089}}{50}}{2\times \frac{1}{4}}$$

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

$$B=\frac{-1±\frac{\sqrt{3089}}{50}}{\frac{1}{2}}$$

Now solve the equation B=\frac{-1±\frac{\sqrt{3089}}{50}}{\frac{1}{2}} when ± is plus. Add -1 to \frac{\sqrt{3089}}{50}.

$$B=\frac{\frac{\sqrt{3089}}{50}-1}{\frac{1}{2}}$$

Divide -1+\frac{\sqrt{3089}}{50} by \frac{1}{2} by multiplying -1+\frac{\sqrt{3089}}{50} by the reciprocal of \frac{1}{2}.

$$B=\frac{\sqrt{3089}}{25}-2$$

Now solve the equation B=\frac{-1±\frac{\sqrt{3089}}{50}}{\frac{1}{2}} when ± is minus. Subtract \frac{\sqrt{3089}}{50} from -1.

$$B=\frac{-\frac{\sqrt{3089}}{50}-1}{\frac{1}{2}}$$

Divide -1-\frac{\sqrt{3089}}{50} by \frac{1}{2} by multiplying -1-\frac{\sqrt{3089}}{50} by the reciprocal of \frac{1}{2}.

$$B=-\frac{\sqrt{3089}}{25}-2$$

The equation is now solved.

$$B=\frac{\sqrt{3089}}{25}-2 B=-\frac{\sqrt{3089}}{25}-2$$

Steps for Completing the Square

Calculate 0.38 to the power of 2 and get 0.1444.

$$0.1444+B+\left(\frac{B}{2}\right)^{2}=0.38$$

To raise \frac{B}{2} to a power, raise both numerator and denominator to the power and then divide.

$$0.1444+B+\frac{B^{2}}{2^{2}}=0.38$$

To add or subtract expressions, expand them to make their denominators the same. Multiply B times \frac{2^{2}}{2^{2}}.

$$0.1444+\frac{B\times 2^{2}}{2^{2}}+\frac{B^{2}}{2^{2}}=0.38$$

Since \frac{B\times 2^{2}}{2^{2}} and \frac{B^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$0.1444+\frac{B\times 2^{2}+B^{2}}{2^{2}}=0.38$$

Do the multiplications in B\times 2^{2}+B^{2}.

$$0.1444+\frac{4B+B^{2}}{2^{2}}=0.38$$

Calculate 2 to the power of 2 and get 4.

$$0.1444+\frac{4B+B^{2}}{4}=0.38$$

Divide each term of 4B+B^{2} by 4 to get B+\frac{1}{4}B^{2}.

$$0.1444+B+\frac{1}{4}B^{2}=0.38$$

Subtract 0.1444 from both sides.

$$B+\frac{1}{4}B^{2}=0.38-0.1444$$

Subtract 0.1444 from 0.38 to get 0.2356.

$$B+\frac{1}{4}B^{2}=0.2356$$

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.

$$\frac{1}{4}B^{2}+B=0.2356$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}B^{2}+B}{\frac{1}{4}}=\frac{0.2356}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$B^{2}+\frac{1}{\frac{1}{4}}B=\frac{0.2356}{\frac{1}{4}}$$

Divide 1 by \frac{1}{4} by multiplying 1 by the reciprocal of \frac{1}{4}.

$$B^{2}+4B=\frac{0.2356}{\frac{1}{4}}$$

Divide 0.2356 by \frac{1}{4} by multiplying 0.2356 by the reciprocal of \frac{1}{4}.

$$B^{2}+4B=\frac{589}{625}$$

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.

$$B^{2}+4B+2^{2}=\frac{589}{625}+2^{2}$$

Square 2.

$$B^{2}+4B+4=\frac{589}{625}+4$$

Add \frac{589}{625} to 4.

$$B^{2}+4B+4=\frac{3089}{625}$$

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}.

$$\left(B+2\right)^{2}=\frac{3089}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(B+2\right)^{2}}=\sqrt{\frac{3089}{625}}$$

Simplify.

$$B+2=\frac{\sqrt{3089}}{25} B+2=-\frac{\sqrt{3089}}{25}$$

Subtract 2 from both sides of the equation.

$$B=\frac{\sqrt{3089}}{25}-2 B=-\frac{\sqrt{3089}}{25}-2$$