Solve -a=sqrt{2a^2+3ab-10}

$$-a = \sqrt{ 2 { a }^{ 2 } +3ab-10 }$$

Solve for b

$$b=-\frac{a}{3}+\frac{10}{3a}$$

$$a<0$$

Steps for Solving Radical Equations

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

$$\sqrt{2a^{2}+3ab-10}=-a$$

Square both sides of the equation.

$$3ab+2a^{2}-10=a^{2}$$

Subtract 2a^{2}-10 from both sides of the equation.

$$3ab+2a^{2}-10-\left(2a^{2}-10\right)=a^{2}-\left(2a^{2}-10\right)$$

Subtracting 2a^{2}-10 from itself leaves 0.

$$3ab=a^{2}-\left(2a^{2}-10\right)$$

Subtract 2a^{2}-10 from a^{2}.

$$3ab=10-a^{2}$$

Divide both sides by 3a.

$$\frac{3ab}{3a}=\frac{10-a^{2}}{3a}$$

Dividing by 3a undoes the multiplication by 3a.

$$b=\frac{10-a^{2}}{3a}$$

Divide -a^{2}+10 by 3a.

$$b=-\frac{a}{3}+\frac{10}{3a}$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{-\sqrt{9b^{2}+40}-3b}{2}\text{, }&arg(\frac{-\sqrt{9b^{2}+40}-3b}{2})\geq \pi \\a=\frac{\sqrt{9b^{2}+40}-3b}{2}\text{, }&arg(\frac{\sqrt{9b^{2}+40}-3b}{2})\geq \pi \end{matrix}\right.$$

Solve for b (complex solution)

$$b=-\frac{a}{3}+\frac{10}{3a}$$

$$arg(a)\geq \pi \text{ and }a\neq 0$$

Solve for a

$$a=\frac{-\sqrt{9b^{2}+40}-3b}{2}$$

$$\left(\frac{\sqrt{9b^{2}+80}}{4}-\frac{\sqrt{9b^{2}+40}}{2}-\frac{3b}{4}\leq 0\text{ or }-\frac{\sqrt{9b^{2}+40}}{2}-\frac{\sqrt{9b^{2}+80}}{4}-\frac{3b}{4}\geq 0\right)\text{ and }b\leq \frac{\sqrt{9b^{2}+40}}{6}+\frac{3b}{2}$$