Solve x=5/2(4+x)

$$x = \frac { 5 } { 2 ( 4 + x ) }$$

Solve for x

$$x=\frac{\sqrt{26}}{2}-2\approx 0.549509757$$

$$x=-\frac{\sqrt{26}}{2}-2\approx -4.549509757$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by 4+x.

$$x=\frac{5}{8+2x}$$

Subtract \frac{5}{8+2x} from both sides.

$$x-\frac{5}{8+2x}=0$$

Factor 8+2x.

$$x-\frac{5}{2\left(x+4\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2\left(x+4\right)}{2\left(x+4\right)}.

$$\frac{x\times 2\left(x+4\right)}{2\left(x+4\right)}-\frac{5}{2\left(x+4\right)}=0$$

Since \frac{x\times 2\left(x+4\right)}{2\left(x+4\right)} and \frac{5}{2\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 2\left(x+4\right)-5}{2\left(x+4\right)}=0$$

Do the multiplications in x\times 2\left(x+4\right)-5.

$$\frac{2x^{2}+8x-5}{2\left(x+4\right)}=0$$

Factor the expressions that are not already factored in \frac{2x^{2}+8x-5}{2\left(x+4\right)}.

$$\frac{2\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)}{2\left(x+4\right)}=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)}{x+4}=0$$

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)=0$$

To find the opposite of -\frac{1}{2}\sqrt{26}-2, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{26}+2\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)=0$$

To find the opposite of \frac{1}{2}\sqrt{26}-2, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{26}+2\right)\left(x-\frac{1}{2}\sqrt{26}+2\right)=0$$

Use the distributive property to multiply x+\frac{1}{2}\sqrt{26}+2 by x-\frac{1}{2}\sqrt{26}+2 and combine like terms.

$$x^{2}+4x-\frac{1}{4}\left(\sqrt{26}\right)^{2}+4=0$$

The square of \sqrt{26} is 26.

$$x^{2}+4x-\frac{1}{4}\times 26+4=0$$

Multiply -\frac{1}{4} and 26 to get -\frac{13}{2}.

$$x^{2}+4x-\frac{13}{2}+4=0$$

Add -\frac{13}{2} and 4 to get -\frac{5}{2}.

$$x^{2}+4x-\frac{5}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -\frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-\frac{5}{2}\right)}}{2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-\frac{5}{2}\right)}}{2}$$

Multiply -4 times -\frac{5}{2}.

$$x=\frac{-4±\sqrt{16+10}}{2}$$

Add 16 to 10.

$$x=\frac{-4±\sqrt{26}}{2}$$

Now solve the equation x=\frac{-4±\sqrt{26}}{2} when ± is plus. Add -4 to \sqrt{26}.

$$x=\frac{\sqrt{26}-4}{2}$$

Divide -4+\sqrt{26} by 2.

$$x=\frac{\sqrt{26}}{2}-2$$

Now solve the equation x=\frac{-4±\sqrt{26}}{2} when ± is minus. Subtract \sqrt{26} from -4.

$$x=\frac{-\sqrt{26}-4}{2}$$

Divide -4-\sqrt{26} by 2.

$$x=-\frac{\sqrt{26}}{2}-2$$

The equation is now solved.

$$x=\frac{\sqrt{26}}{2}-2 x=-\frac{\sqrt{26}}{2}-2$$

Steps for Completing the Square

Use the distributive property to multiply 2 by 4+x.

$$x=\frac{5}{8+2x}$$

Subtract \frac{5}{8+2x} from both sides.

$$x-\frac{5}{8+2x}=0$$

Factor 8+2x.

$$x-\frac{5}{2\left(x+4\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2\left(x+4\right)}{2\left(x+4\right)}.

$$\frac{x\times 2\left(x+4\right)}{2\left(x+4\right)}-\frac{5}{2\left(x+4\right)}=0$$

Since \frac{x\times 2\left(x+4\right)}{2\left(x+4\right)} and \frac{5}{2\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 2\left(x+4\right)-5}{2\left(x+4\right)}=0$$

Do the multiplications in x\times 2\left(x+4\right)-5.

$$\frac{2x^{2}+8x-5}{2\left(x+4\right)}=0$$

Factor the expressions that are not already factored in \frac{2x^{2}+8x-5}{2\left(x+4\right)}.

$$\frac{2\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)}{2\left(x+4\right)}=0$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)}{x+4}=0$$

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

$$\left(x-\left(-\frac{1}{2}\sqrt{26}-2\right)\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)=0$$

To find the opposite of -\frac{1}{2}\sqrt{26}-2, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{26}+2\right)\left(x-\left(\frac{1}{2}\sqrt{26}-2\right)\right)=0$$

To find the opposite of \frac{1}{2}\sqrt{26}-2, find the opposite of each term.

$$\left(x+\frac{1}{2}\sqrt{26}+2\right)\left(x-\frac{1}{2}\sqrt{26}+2\right)=0$$

Use the distributive property to multiply x+\frac{1}{2}\sqrt{26}+2 by x-\frac{1}{2}\sqrt{26}+2 and combine like terms.

$$x^{2}+4x-\frac{1}{4}\left(\sqrt{26}\right)^{2}+4=0$$

The square of \sqrt{26} is 26.

$$x^{2}+4x-\frac{1}{4}\times 26+4=0$$

Multiply -\frac{1}{4} and 26 to get -\frac{13}{2}.

$$x^{2}+4x-\frac{13}{2}+4=0$$

Add -\frac{13}{2} and 4 to get -\frac{5}{2}.

$$x^{2}+4x-\frac{5}{2}=0$$

Add \frac{5}{2} to both sides. Anything plus zero gives itself.

$$x^{2}+4x=\frac{5}{2}$$

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.

$$x^{2}+4x+2^{2}=\frac{5}{2}+2^{2}$$

Square 2.

$$x^{2}+4x+4=\frac{5}{2}+4$$

Add \frac{5}{2} to 4.

$$x^{2}+4x+4=\frac{13}{2}$$

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

$$\left(x+2\right)^{2}=\frac{13}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{13}{2}}$$

Simplify.

$$x+2=\frac{\sqrt{26}}{2} x+2=-\frac{\sqrt{26}}{2}$$

Subtract 2 from both sides of the equation.

$$x=\frac{\sqrt{26}}{2}-2 x=-\frac{\sqrt{26}}{2}-2$$