## Precalculus (6th Edition)

The solutions are $x=1\pm\dfrac{\sqrt{10}}{2}$
$-2x^{2}+4x+3=0$ 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 $x$: $x=1\pm\dfrac{\sqrt{10}}{2}$