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Solve for x: x²+4x=5x+12

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\begin{gathered} x^2+4x=5x+12 \\ x^2+4x-12=5x+12-12 \\ x^2+4x-12=5x \\ x^2+4x-12-5x=5x-5x \\ x^2-x-12=0 \end{gathered}

then factoring the equation


\begin{gathered} x^2+3x-4x-12=0 \\ (x^2+3x)+(-4x-12)=0 \\ x(x+3)-4(x+3)=0 \\ (x+3)(x-4)=0 \\ x+3=0\quad \mathrm{or}\quad \: x-4=0 \end{gathered}

so


\begin{gathered} x+3=0 \\ x+3-3=0-3 \\ x=-3 \\ \text{and} \\ x-4=0 \\ x-4+4=0+4 \\ x=4 \end{gathered}

solution:

x = -3

x = 4

User Rodrigo Lira
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