Solve 1/3u^2+3/2u=-1/2b

$$\frac { 1 } { 3 } u ^ { 2 } + \frac { 3 } { 2 } u = - \frac { 1 } { 2 } b$$

Solve for b

$$b=\frac{u\left(-2u-9\right)}{3}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$-\frac{1}{2}b=\frac{1}{3}u^{2}+\frac{3}{2}u$$

The equation is in standard form.

$$-\frac{1}{2}b=\frac{u^{2}}{3}+\frac{3u}{2}$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}b}{-\frac{1}{2}}=\frac{\frac{u^{2}}{3}+\frac{3u}{2}}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$b=\frac{\frac{u^{2}}{3}+\frac{3u}{2}}{-\frac{1}{2}}$$

Divide \frac{3u}{2}+\frac{u^{2}}{3} by -\frac{1}{2} by multiplying \frac{3u}{2}+\frac{u^{2}}{3} by the reciprocal of -\frac{1}{2}.

$$b=-\frac{2u^{2}}{3}-3u$$

Solve for u (complex solution)

$$u=\frac{\sqrt{81-24b}-9}{4}$$

$$u=\frac{-\sqrt{81-24b}-9}{4}$$

Solve for u

$$u=\frac{\sqrt{81-24b}-9}{4}$$

$$u=\frac{-\sqrt{81-24b}-9}{4}\text{, }b\leq \frac{27}{8}$$