Solve n(6x-5)^2-(36x^2+4)=-(2x+5)-32

$$n ( 6 x - 5 ) ^ { 2 } - ( 36 x ^ { 2 } + 4 ) = - ( 2 x + 5 ) - 32$$

Solve for n

$$n=-\frac{33+2x-36x^{2}}{\left(6x-5\right)^{2}}$$

$$x\neq \frac{5}{6}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-5\right)^{2}.

$$n\left(36x^{2}-60x+25\right)-\left(36x^{2}+4\right)=-\left(2x+5\right)-32$$

Use the distributive property to multiply n by 36x^{2}-60x+25.

$$36nx^{2}-60nx+25n-\left(36x^{2}+4\right)=-\left(2x+5\right)-32$$

To find the opposite of 36x^{2}+4, find the opposite of each term.

$$36nx^{2}-60nx+25n-36x^{2}-4=-\left(2x+5\right)-32$$

To find the opposite of 2x+5, find the opposite of each term.

$$36nx^{2}-60nx+25n-36x^{2}-4=-2x-5-32$$

Subtract 32 from -5 to get -37.

$$36nx^{2}-60nx+25n-36x^{2}-4=-2x-37$$

Add 36x^{2} to both sides.

$$36nx^{2}-60nx+25n-4=-2x-37+36x^{2}$$

Add 4 to both sides.

$$36nx^{2}-60nx+25n=-2x-37+36x^{2}+4$$

Add -37 and 4 to get -33.

$$36nx^{2}-60nx+25n=-2x-33+36x^{2}$$

Combine all terms containing n.

$$\left(36x^{2}-60x+25\right)n=-2x-33+36x^{2}$$

The equation is in standard form.

$$\left(36x^{2}-60x+25\right)n=36x^{2}-2x-33$$

Divide both sides by 36x^{2}-60x+25.

$$\frac{\left(36x^{2}-60x+25\right)n}{36x^{2}-60x+25}=\frac{36x^{2}-2x-33}{36x^{2}-60x+25}$$

Dividing by 36x^{2}-60x+25 undoes the multiplication by 36x^{2}-60x+25.

$$n=\frac{36x^{2}-2x-33}{36x^{2}-60x+25}$$

Divide -2x-33+36x^{2} by 36x^{2}-60x+25.

$$n=\frac{36x^{2}-2x-33}{\left(6x-5\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{30n+\sqrt{29\left(41-12n\right)}-1}{36\left(n-1\right)}\text{; }x=\frac{30n-\sqrt{29\left(41-12n\right)}-1}{36\left(n-1\right)}\text{, }&n\neq 1\\x=1\text{, }&n=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{30n+\sqrt{29\left(41-12n\right)}-1}{36\left(n-1\right)}\text{; }x=\frac{30n-\sqrt{29\left(41-12n\right)}-1}{36\left(n-1\right)}\text{, }&n\neq 1\text{ and }n\leq \frac{41}{12}\\x=1\text{, }&n=1\end{matrix}\right.$$