Solve 2BC^2+10sqrt{2}BC+14=0

$$2 B C ^ { 2 } + 10 \sqrt { 2 } B C + 14 = 0$$

Solve for B

$$B=-\frac{7}{C\left(C+5\sqrt{2}\right)}$$

$$C\neq -5\sqrt{2}\text{ and }C\neq 0$$

Steps for Solving Linear Equation

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$2BC^{2}+10\sqrt{2}BC=-14$$

Combine all terms containing B.

$$\left(2C^{2}+10\sqrt{2}C\right)B=-14$$

Divide both sides by 2C^{2}+10\sqrt{2}C.

$$\frac{\left(2C^{2}+10\sqrt{2}C\right)B}{2C^{2}+10\sqrt{2}C}=-\frac{14}{2C^{2}+10\sqrt{2}C}$$

Dividing by 2C^{2}+10\sqrt{2}C undoes the multiplication by 2C^{2}+10\sqrt{2}C.

$$B=-\frac{14}{2C^{2}+10\sqrt{2}C}$$

Divide -14 by 2C^{2}+10\sqrt{2}C.

$$B=-\frac{7}{C\left(C+5\sqrt{2}\right)}$$

Solve for C (complex solution)

$$C=-\frac{\sqrt{2}\left(\sqrt{B\left(25B-14\right)}+5B\right)}{2B}$$

$$C=\frac{\sqrt{2}\left(\sqrt{B\left(25B-14\right)}-5B\right)}{2B}\text{, }B\neq 0$$

Solve for C

$$\left\{\begin{matrix}C=\frac{-\sqrt{50-\frac{28}{B}}-5\sqrt{2}}{2}\text{; }C=\frac{\sqrt{50-\frac{28}{B}}-5\sqrt{2}}{2}\text{, }&B\geq \frac{14}{25}\\C=-\frac{\sqrt{50B^{2}-28B}}{2B}-\frac{5\sqrt{2}}{2}\text{; }C=\frac{\sqrt{50B^{2}-28B}}{2B}-\frac{5\sqrt{2}}{2}\text{, }&B<0\end{matrix}\right.$$