Answer
{$\dfrac{3 - \sqrt {65}}{4},\dfrac{3 + \sqrt {65}}{4}$}
Work Step by Step
Given: $(2x-5)(x+1)=0$
Re-write the given equation as: $2x^2-3x-7=0$
Factorize the expression with the help of quadratic formula. Quadratic formula suggests that $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
This implies that $x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-7)}}{2(2)}$
or, $x=\dfrac{3 \pm \sqrt {65}}{4}$
or, $x=\dfrac{3 - \sqrt {65}}{4},\dfrac{3 + \sqrt {65}}{4}$
Hence, our solution set is: {$\dfrac{3 - \sqrt {65}}{4},\dfrac{3 + \sqrt {65}}{4}$}
