10x+20=2x^2

Solution for 10x+20=2x^2 equation:

10x+20=2x^2

We move all terms to the left:

10x+20-(2x^2)=0

determiningTheFunctionDomain
-2x^2+10x+20=0

a = -2; b = 10; c = +20;
Δ = b2-4ac
Δ = 102-4·(-2)·20
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{65}}{2*-2}=\frac{-10-2\sqrt{65}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{65}}{2*-2}=\frac{-10+2\sqrt{65}}{-4}
$