Solve n3(y+1)^2-10(y+1)+8=0

$$n 3 ( y + 1 ) ^ { 2 } - 10 ( y + 1 ) + 8 = 0$$

Solve for n_3

$$n_{3}=\frac{2\left(5y+1\right)}{\left(y+1\right)^{2}}$$

$$y\neq -1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$n_{3}\left(y^{2}+2y+1\right)-10\left(y+1\right)+8=0$$

Use the distributive property to multiply n_{3} by y^{2}+2y+1.

$$n_{3}y^{2}+2n_{3}y+n_{3}-10\left(y+1\right)+8=0$$

Use the distributive property to multiply -10 by y+1.

$$n_{3}y^{2}+2n_{3}y+n_{3}-10y-10+8=0$$

Add -10 and 8 to get -2.

$$n_{3}y^{2}+2n_{3}y+n_{3}-10y-2=0$$

Add 10y to both sides. Anything plus zero gives itself.

$$n_{3}y^{2}+2n_{3}y+n_{3}-2=10y$$

Add 2 to both sides.

$$n_{3}y^{2}+2n_{3}y+n_{3}=10y+2$$

Combine all terms containing n_{3}.

$$\left(y^{2}+2y+1\right)n_{3}=10y+2$$

Divide both sides by y^{2}+2y+1.

$$\frac{\left(y^{2}+2y+1\right)n_{3}}{y^{2}+2y+1}=\frac{10y+2}{y^{2}+2y+1}$$

Dividing by y^{2}+2y+1 undoes the multiplication by y^{2}+2y+1.

$$n_{3}=\frac{10y+2}{y^{2}+2y+1}$$

Divide 10y+2 by y^{2}+2y+1.

$$n_{3}=\frac{2\left(5y+1\right)}{\left(y+1\right)^{2}}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{n_{3}+\sqrt{25-8n_{3}}-5}{n_{3}}\text{; }y=-\frac{n_{3}-\sqrt{25-8n_{3}}-5}{n_{3}}\text{, }&n_{3}\neq 0\\y=-\frac{1}{5}\text{, }&n_{3}=0\end{matrix}\right.$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{n_{3}+\sqrt{25-8n_{3}}-5}{n_{3}}\text{; }y=-\frac{n_{3}-\sqrt{25-8n_{3}}-5}{n_{3}}\text{, }&n_{3}\neq 0\text{ and }n_{3}\leq \frac{25}{8}\\y=-\frac{1}{5}\text{, }&n_{3}=0\end{matrix}\right.$$