A standard quadratic equation is given as
\[ax^2+bx+c=0\]
its roots or its solution is given by the following formula
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
It's very easy, I'll show you
your question is
-5+2x^2=6x
The first step is to bring all the terms to one side
\[-5+2x^2=6x\]
Let's subtract both sides by 6x
we get
\[-5+2x^2-6x=6x-6x\]
we'll get
\[-5+2x^2-6x=0\]
Now let's write this in descending powers of x
\[2x^2-6x-5\]
Did you understand till now?
Now its solution is given by
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
Let's substitute a=2, b=-6 and c=-5
\[x=\frac{-(-6)\pm \sqrt{(-6)^2-4\times (2)\times (-5)}}{2\times (2)}\]
We get now
\[x=\frac{6\pm \sqrt{36-(-40)}}{4}\]
\[x=\frac{6\pm \sqrt{36+40}}{4}\]
\[x=\frac{6\pm \sqrt{76}}{4}\]
Did you understand this?
We've almost found the solution, we could simplify it more
We have
\[x=\frac{6\pm \sqrt{76}}{4}\]
\[\sqrt {76}= \sqrt{4\times 19}=2\sqrt {19}\]
so
\[x=\frac{6\pm 2\sqrt{19}}{4}\]
divide the whole equation by 2
we get
\[x=\frac{3\pm \sqrt {19}}{2}\]
so the two solutions\roots are
\[x=\frac{3+ \sqrt {19}}{2}, \frac{3- \sqrt {19}}{2}\]\]
Did you understand?