Solve 2(1/2)^3+(1/2)^2p-25(1/2)+4=0

$$2 ( \frac { 1 } { 2 } ) ^ { 3 } + ( \frac { 1 } { 2 } ) ^ { 2 } p - 25 ( \frac { 1 } { 2 } ) + 4 = 0$$

Solve for p

$$p=33$$

Steps for Solving Linear Equation

Calculate \frac{1}{2} to the power of 3 and get \frac{1}{8}.

$$2\times \frac{1}{8}+\left(\frac{1}{2}\right)^{2}p-25\times \frac{1}{2}+4=0$$

Multiply 2 and \frac{1}{8} to get \frac{2}{8}.

$$\frac{2}{8}+\left(\frac{1}{2}\right)^{2}p-25\times \frac{1}{2}+4=0$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$\frac{1}{4}+\left(\frac{1}{2}\right)^{2}p-25\times \frac{1}{2}+4=0$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\frac{1}{4}+\frac{1}{4}p-25\times \frac{1}{2}+4=0$$

Multiply 25 and \frac{1}{2} to get \frac{25}{2}.

$$\frac{1}{4}+\frac{1}{4}p-\frac{25}{2}+4=0$$

Least common multiple of 4 and 2 is 4. Convert \frac{1}{4} and \frac{25}{2} to fractions with denominator 4.

$$\frac{1}{4}+\frac{1}{4}p-\frac{50}{4}+4=0$$

Since \frac{1}{4} and \frac{50}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-50}{4}+\frac{1}{4}p+4=0$$

Subtract 50 from 1 to get -49.

$$-\frac{49}{4}+\frac{1}{4}p+4=0$$

Convert 4 to fraction \frac{16}{4}.

$$-\frac{49}{4}+\frac{1}{4}p+\frac{16}{4}=0$$

Since -\frac{49}{4} and \frac{16}{4} have the same denominator, add them by adding their numerators.

$$\frac{-49+16}{4}+\frac{1}{4}p=0$$

Add -49 and 16 to get -33.

$$-\frac{33}{4}+\frac{1}{4}p=0$$

Add \frac{33}{4} to both sides. Anything plus zero gives itself.

$$\frac{1}{4}p=\frac{33}{4}$$

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$p=\frac{33}{4}\times 4$$

Cancel out 4 and 4.

$$p=33$$