Answer
the solution set is\[\left\{ 1,\,\,\frac{5}{7} \right\}\].
Work Step by Step
The given equation can be written as\[7x\left( x-2 \right)=3-2\left( x+4 \right)\].
Apply distributive law on both the sides as follows:
\[\begin{align}
& 7x\left( x-2 \right)=3-2\left( x+4 \right) \\
& 7{{x}^{2}}-14=3-2x-8 \\
& 7{{x}^{2}}-14=-2x-5 \\
\end{align}\]
Simplifying
\[7{{x}^{2}}-12x+5=0\]
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\], where\[a=7,\ b=-12,\text{ and }c=5\].
Now, put these values in the quadratic formula \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\begin{align}
& x=\frac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\times \left( 7 \right)\times \left( 5 \right)}}{2\times 7} \\
& =\frac{12\pm \sqrt{144-140}}{4} \\
& =\frac{-12\pm \sqrt{4}}{4} \\
& =\frac{-12\pm 2}{4}
\end{align}\]
This gives,\[x=1,\,\,\frac{5}{7}\]
Hence, the solution set is\[\left\{ 1,\,\,\frac{5}{7} \right\}\].
