## College Algebra (10th Edition)

The solution set is $\left\{-\frac{5}{2}, \frac{4}{3}\right\}$.
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 $\left\{-\frac{5}{2}, \frac{4}{3}\right\}$.