Answer
$x=\dfrac{1-\sqrt{7}}{2}
\text{ and }
x=\dfrac{1+\sqrt{7}}{2}
$
Work Step by Step
Using the properties of equality, the given equation, $
2x(x-1)=3
,$ is equivalent to
\begin{align*}
2x(x)+2x(-1)&=3
&\text{ (use Distributive Property)}
\\
2x^2-2x&=3
\\
2x^2-2x-3&=0
\end{align*}
In the equation above, $a=
2
,$ $b=
-2
,$ and $c=
-3
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}\require{cancel}x&=
\dfrac{-(-2)\pm\sqrt{(-2)^2-4(2)(-3)}}{2(2)}
\\\\&=
\dfrac{2\pm\sqrt{4+24}}{4}
\\\\&=
\dfrac{2\pm\sqrt{28}}{4}
\\\\&=
\dfrac{2\pm\sqrt{4\cdot7}}{4}
\\\\&=
\dfrac{2\pm2\sqrt{7}}{4}
\\\\&=
\dfrac{\cancel2^1\pm\cancel2^1\sqrt{7}}{\cancel4^2}
&\text{ (divide by $2$)}
\\\\&=
\dfrac{1\pm\sqrt{7}}{2}
\end{align*}
Hence, the solutions are$
x=\dfrac{1-\sqrt{7}}{2}
\text{ and }
x=\dfrac{1+\sqrt{7}}{2}
.$
