How to solve this equation?

Solve:

#5(x+1)^2+7(x+3)^2=12(x+2)^2#

Expand using the sum of squares: #(a+b)^2=a^2+2ab+b^2#, or using the FOIL method: https://www.ipracticemath.com/learn/algebra/foil-method-of-binomial-multiplication

#5(x^2+2x+1)+7(x^2+6x+9)=12(x^2+4x+4)#

Expand using the distributive property: #a(b+c)=ab+ac#

#5x^2+10x+5+7x^2+42x+63=12x^2+48x+48#

Collect like terms.

#(5x^2+7x^2)+(10x+42x)+(5+63)=12x^2+48x+48#

Combine like terms.

#12x^2+52x+68=12x^2+48x+48#

Cancel #12x^2# on both sides.

#color(red)cancel(color(black)(12x^2))+52x+68=color(red)cancel(color(black)(12x^2))+48x+48#

Simplify.

#52x+68=48x+48#

Subtract #48x# from both sides.

#52x-48x+68=48#

Simplify.

#4x+68=48#

Subtract #68# from both sides.

#4x=48-68#

Simplify.

#4x=-20#

Divide both sides by #4#.

#x=-20/4#

Simplify.

#x=-5#

Step by step, performing legal algebraic operations that just happen to simplify the expression:

#5(x+1)^2 + 7(x+3)^2 = 12(x+2)^2#

#5(x^2+2x+1) + 7(x^2+6x+9) = 12(x^2+4x+4)#

#(5x^2+10x+5) + (7x^2+42x+63) = (12x^2+48x+48)#

#(12x^2+52x+68) = (12x^2+48x+48)#

#(12x^2+52x+68) - (12x^2+48x+48) = 0#

#(0x^2+4x+20) = 0#

#4x+20 = 0#

#4x = -20#

#1/cancel(4) * cancel(4)x = 1/cancel(4) * (-cancel(20)^5)#

#x = -5#

I hope this helps,
Steve