Answer
$\left \{\frac{-2 \pm \sqrt {14}}{2}\right\}$.
Work Step by Step
Add $-5+4x$ to each side of the given equation.
$2x^2-5+4x=5-4x-5+4x$
Simplify.
$2x^2+4x-5=0$ ...... (1)
The standard form of the quadratic equation is
$ax^2+bx+c=0$
and the solutions can be calculated using the Quadratic Formula: $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
Comparing equation (1) and the standard equation we identify $a$, $b$ , $c$:
$a=2,b=4$ and $c=-5$
Hence, the solutions are:
$x=\frac{-4\pm \sqrt {4^2-4(2)(-5)}}{2(2)}$
Simplify.
$x=\frac{-4 \pm \sqrt {16+40}}{4}$
$x=\frac{-4 \pm \sqrt {56}}{4}$
$x=\frac{-4 \pm \sqrt {2^2\cdot 14}}{4}$
$x=\frac{-4 \pm 2\sqrt {14}}{4}$
$x=\frac{-2 \pm \sqrt {14}}{2}$
The solution set is $\left \{\frac{-2 \pm \sqrt {14}}{2}\right\}$.
