Answer
The solution set is $\left\{-\frac{5}{2}, \frac{4}{3}\right\}$.
Work Step by Step
RECALL:
A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula
$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
The given quadratic equation has:
$a=6
\\b=7
\\c=-20$
Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-7 \pm \sqrt{7^2-4(6)(-20)}}{2(6)}
\\x=\dfrac{-7 \pm \sqrt{49-(-480)}}{12}
\\x=\dfrac{-7\pm \sqrt{49+480}}{12}
\\x =\dfrac{-7\pm\sqrt{529}}{12}
\\x=\dfrac{-7\pm\sqrt{23^2}}{12}
\\x=\dfrac{-7\pm 23}{12}$
Split the solutions to obtain:
$x_1 = \dfrac{-7+23}{12} = \dfrac{16}{12} = \dfrac{4}{3}
\\x_2=\dfrac{-7-23}{12}=\dfrac{-30}{12}=-\dfrac{5}{2}$
Therefore, the solution set is $\left\{-\frac{5}{2}, \frac{4}{3}\right\}$.
