Solve [(2m-n)^2+(m-2n)(m+2n)-5m(m+n)]=(-3n)

$$[ ( 2 m - n ) ^ { 2 } + ( m - 2 n ) ( m + 2 n ) - 5 m ( m + n ) ] = ( - 3 n )$$

Solve for m

$$\left\{\begin{matrix}\\m=\frac{1-n}{3}\text{, }&\text{unconditionally}\\m\in \mathrm{R}\text{, }&n=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$4m^{2}-4mn+n^{2}+\left(m-2n\right)\left(m+2n\right)-5m\left(m+n\right)=-3n$$

Consider \left(m-2n\right)\left(m+2n\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4m^{2}-4mn+n^{2}+m^{2}-\left(2n\right)^{2}-5m\left(m+n\right)=-3n$$

Expand \left(2n\right)^{2}.

$$4m^{2}-4mn+n^{2}+m^{2}-2^{2}n^{2}-5m\left(m+n\right)=-3n$$

Calculate 2 to the power of 2 and get 4.

$$4m^{2}-4mn+n^{2}+m^{2}-4n^{2}-5m\left(m+n\right)=-3n$$

Combine 4m^{2} and m^{2} to get 5m^{2}.

$$5m^{2}-4mn+n^{2}-4n^{2}-5m\left(m+n\right)=-3n$$

Combine n^{2} and -4n^{2} to get -3n^{2}.

$$5m^{2}-4mn-3n^{2}-5m\left(m+n\right)=-3n$$

Use the distributive property to multiply -5m by m+n.

$$5m^{2}-4mn-3n^{2}-5m^{2}-5mn=-3n$$

Combine 5m^{2} and -5m^{2} to get 0.

$$-4mn-3n^{2}-5mn=-3n$$

Combine -4mn and -5mn to get -9mn.

$$-9mn-3n^{2}=-3n$$

Add 3n^{2} to both sides.

$$-9mn=-3n+3n^{2}$$

The equation is in standard form.

$$\left(-9n\right)m=3n^{2}-3n$$

Divide both sides by -9n.

$$\frac{\left(-9n\right)m}{-9n}=\frac{3n\left(n-1\right)}{-9n}$$

Dividing by -9n undoes the multiplication by -9n.

$$m=\frac{3n\left(n-1\right)}{-9n}$$

Divide 3n\left(-1+n\right) by -9n.

$$m=\frac{1-n}{3}$$

Solve for n

$$n=1-3m$$

$$n=0$$