Answer
$\left\{\dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2}\right\}$
Work Step by Step
We have to solve the equation:
$$2x^2=3-4x.$$
Bring the equation to the standard form by moving all terms on one side:
$$\begin{align*}
2x^2+4x-3&=0.
\end{align*}$$
To solve the equation $ax^2+bx+c=0$ we will use the quadratic formula:
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Identify $a$, $b$, $c$:
$$\begin{align*}
a&=2\\
b&=4\\
c&=-3.
\end{align*}$$
We solve the given equation by substituting the values of $a$, $b$, $c$ in the quadratic formula:
$$\begin{align*}
x&=\dfrac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)}\\
&=\dfrac{-4\pm\sqrt{40}}{4}\\
&=\dfrac{-4\pm 2\sqrt {10}}{4}\\
&=\dfrac{-2\pm\sqrt{10}}{2}.
\end{align*}$$
The solution set of the equation is:
$$\left\{\dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2}\right\}.$$
