Answer
{$\dfrac{-3 - \sqrt {17}}{4},\dfrac{-3 +\sqrt {17}}{4}$}
Work Step by Step
Re-write the given equation as: $2x^2+3x-1=0$
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 $x=\dfrac{-(3) \pm \sqrt{(3)^2-4(2)(-1)}}{2(2)}$
or, $x=\dfrac{-3 \pm \sqrt {9+8}}{4}$
or, $x=\dfrac{-3 \pm \sqrt {17}}{4}$
or, $x=\dfrac{-3 - \sqrt {17}}{4},\dfrac{-3 +\sqrt {17}}{4}$
Hence, our solution set is: {$\dfrac{-3 - \sqrt {17}}{4},\dfrac{-3 +\sqrt {17}}{4}$}
