Solve 3a^3d+14x^2+7x=25

$$3 a ^ { 3 } d + 14 x ^ { 2 } + 7 x = 25$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{14x^{2}+7x-25}{3a^{3}}\text{, }&a\neq 0\\d\in \mathrm{R}\text{, }&\left(x=-\frac{3\sqrt{161}}{28}-\frac{1}{4}\text{ or }x=\frac{3\sqrt{161}}{28}-\frac{1}{4}\right)\text{ and }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 14x^{2} from both sides.

$$3a^{3}d+7x=25-14x^{2}$$

Subtract 7x from both sides.

$$3a^{3}d=25-14x^{2}-7x$$

The equation is in standard form.

$$3a^{3}d=25-7x-14x^{2}$$

Divide both sides by 3a^{3}.

$$\frac{3a^{3}d}{3a^{3}}=\frac{25-7x-14x^{2}}{3a^{3}}$$

Dividing by 3a^{3} undoes the multiplication by 3a^{3}.

$$d=\frac{25-7x-14x^{2}}{3a^{3}}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{3^{\frac{2}{3}}\sqrt[3]{\frac{14x^{2}+7x-25}{d}}}{3}\text{, }&d\neq 0\\a\in \mathrm{R}\text{, }&\left(x=-\frac{3\sqrt{161}}{28}-\frac{1}{4}\text{ or }x=\frac{3\sqrt{161}}{28}-\frac{1}{4}\right)\text{ and }d=0\end{matrix}\right.$$