Answer
The solution set is $\left\{-\frac{3}{5}, \frac{5}{2}\right\}$.
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:
$a=10
\\b=-19
\\c=-15$
Substitute these values into the quadratic formula to obtain:
$x=\dfrac{-(-19) \pm \sqrt{(-19)^2-4(10)(-15)}}{2(10)}
\\x=\dfrac{19 \pm \sqrt{361-(-600)}}{20}
\\x=\dfrac{19\pm \sqrt{361+600}}{20}
\\x =\dfrac{19\pm\sqrt{961}}{20}
\\x=\dfrac{19\pm\sqrt{31^2}}{20}
\\x=\dfrac{19\pm31}{20}$
Split the solutions to obtain:
$x_1 = \dfrac{19+31}{20} = \dfrac{50}{20} = \dfrac{5}{2}
\\x_2=\dfrac{19-31}{20}=\dfrac{-12}{20}=-\dfrac{3}{5}$
Therefore, the solution set is $\left\{-\frac{3}{5}, \frac{5}{2}\right\}$.
