2x^2=3+2x

Solution for 2x^2=3+2x equation:

2x^2=3+2x

We move all terms to the left:

2x^2-(3+2x)=0

We add all the numbers together, and all the variables

2x^2-(2x+3)=0

We get rid of parentheses

2x^2-2x-3=0

a = 2; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·2·(-3)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*2}=\frac{2-2\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*2}=\frac{2+2\sqrt{7}}{4}
$