Answer
$\left\{\dfrac{-7\pm\sqrt{37}}{2}\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
x(2x-7)=3x^2+3
,$ is equivalent to
\begin{align*}
x(2x)+x(-7)&=3x^2+3
&(\text{use Distributive Property}
\\
2x^2-7x&=3x^2+3
\\
0&=(3x^2-2x^2)+7x+3
\\
0&=x^2+7x+3
\\
x^2+7x+3&=0
.\end{align*}
The equation above has
\begin{align*}a=
1
,b=
7
,\text{ and }c=
3
.\end{align*}
Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}
x&=\dfrac{-7\pm\sqrt{7^2-4(1)(3)}}{2(1)}
\\\\&=
\dfrac{-7\pm\sqrt{49-12}}{2}
\\\\&=
\dfrac{-7\pm\sqrt{37}}{2}
.\end{align*}
Hence, the solution set of $
x(2x-7)=3x^2+3
$ is $
\left\{\dfrac{-7\pm\sqrt{37}}{2}\right\}
$.
