Answer
The solutions are $x=-\dfrac{3}{4}\pm\dfrac{\sqrt{17}}{4}$
Work Step by Step
$2x^{2}+3x=1$
Take the $1$ to the left side of the equation:
$2x^{2}+3x-1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
In this case, $a=2$, $b=3$ and $c=-1$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(2)(-1)}}{2(2)}=\dfrac{-3\pm\sqrt{9+8}}{4}=...$
$...=\dfrac{-3\pm\sqrt{17}}{4}=-\dfrac{3}{4}\pm\dfrac{\sqrt{17}}{4}$
