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Solve each of the following equation

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(2x+3)^2=(3x-1)^2

start by expanding both sides of the equation


\begin{gathered} (2x)^2+2\cdot2x\cdot3+3^2=(3x)^2-2\cdot3x\cdot1+1^2 \\ 4x^2+12x+9=9x^2-6x+1 \end{gathered}

bring all to the left side of the equation


\begin{gathered} 4x^2+12x+9-9x^2+6x-1=0 \\ -5x^2+18x+8=0 \end{gathered}

switch the signs of the equations


5x^2-18x-8=0

write -18x as a difference


5x^2+2x-20x-8=0

factor x and factor -4


x\cdot(5x+2)-4\cdot(5x+2)=0

factor '5x+2' of the expression


(5x+2)\cdot(x-4)=0

when a product of two factors is equal to 0, at least one of the factors is 0, then equal each to 0 and solve for x


\begin{gathered} 5x+2=0 \\ 5x=-2 \\ x=-(2)/(5) \\ \\ x-4=0 \\ x=4 \end{gathered}

the solutions to the equation are 4 and -2/5.

User Golddove
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