Solve 9g^2+30g+25=(3g+[1])^2

$$9 g ^ { 2 } + 30 g + 25 = ( 3 g + [ 1 ] ) ^ { 2 }$$

Solve for g

$$g=-1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3g+1\right)^{2}.

$$9g^{2}+30g+25=9g^{2}+6g+1$$

Subtract 9g^{2} from both sides.

$$9g^{2}+30g+25-9g^{2}=6g+1$$

Combine 9g^{2} and -9g^{2} to get 0.

$$30g+25=6g+1$$

Subtract 6g from both sides.

$$30g+25-6g=1$$

Combine 30g and -6g to get 24g.

$$24g+25=1$$

Subtract 25 from both sides.

$$24g=1-25$$

Subtract 25 from 1 to get -24.

$$24g=-24$$

Divide both sides by 24.

$$g=\frac{-24}{24}$$

Divide -24 by 24 to get -1.

$$g=-1$$