Answer
The solution set is\[\left\{ 1,\,\,\frac{5}{2} \right\}\].
Work Step by Step
The given equation can be written as\[\left( 2x-6 \right)\left( x+2 \right)=5\left( x-1 \right)-12\].
Use the FOIL method to multiply the terms written on the left sides
\[\begin{align}
& \left( 2x-6 \right)\left( x+2 \right)=\underbrace{2x\cdot x}_{\text{F}}+\underbrace{2x\cdot 2}_{\text{O}}+\underbrace{\left( -6 \right)\cdot x}_{\text{I}}+\underbrace{\left( -6 \right)\cdot 2}_{\text{L}} \\
& =2{{x}^{2}}+4x-6x-12 \\
& =2{{x}^{2}}-2x-12
\end{align}\]
where, “F” stands for the product of the first term, “O” stands for the product of the outside terms, “I” stands for the product of the inside term, and “L” stands for the product of the last term.
So,
\[\left( 2x-6 \right)\left( x+2 \right)=2{{x}^{2}}-2x-12\]
Therefore,
\[\begin{align}
& \left( 2x-6 \right)\left( x+2 \right)=5\left( x-1 \right)-12 \\
& 2{{x}^{2}}-2x-12=5x-5-12 \\
& 2{{x}^{2}}-2x-12=5x-17 \\
\end{align}\]
Simplifying
\[2{{x}^{2}}-7x+5=0\]
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\], where\[a=2,\ b=-7,\ \text{and }c=5\].
Now, put these values in the quadratic formula \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
& x=\frac{-\left( -7 \right)\pm \sqrt{{{\left( -7 \right)}^{2}}-4\times \left( 2 \right)\times \left( 5 \right)}}{2\times 2} \\
& =\frac{7\pm \sqrt{49-40}}{4} \\
& =\frac{-7\pm \sqrt{9}}{4} \\
& =\frac{-7\pm 3}{4}
\end{align}\]
This gives,\[x=1,\,\,\frac{5}{2}\]
Hence, the solution set is\[\left\{ 1,\,\,\frac{5}{2} \right\}\].
