#### Answer

The solutions are $x=1\pm\dfrac{\sqrt{10}}{2}$

#### Work Step by Step

$-2x^{2}+4x+3=0$
Divide the whole equation by $-2$
$-\dfrac{1}{2}(-2x^{2}+4x+3=0)$
$x^{2}-2x-\dfrac{3}{2}=0$
Take $\dfrac{3}{2}$ to the right side:
$x^{2}-2x=\dfrac{3}{2}$
Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-2$
$x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}=\dfrac{3}{2}+\Big(-\dfrac{2}{2}\Big)^{2}$
$x^{2}-2x+1=\dfrac{3}{2}+1$
$x^{2}-2x+1=\dfrac{5}{2}$
Factor the left side of the equation, which is a perfect square trinomial:
$(x-1)^{2}=\dfrac{5}{2}$
Take the square root of both sides:
$\sqrt{(x-1)^{2}}=\pm\sqrt{\dfrac{5}{2}}$
$x-1=\pm\dfrac{\sqrt{10}}{2}$
Solve for $x$:
$x=1\pm\dfrac{\sqrt{10}}{2}$