3x^2=4x=2x^2+3

Solution for 3x^2=4x=2x^2+3 equation:

3x^2=4x=2x^2+3

We move all terms to the left:

3x^2-(4x)=0

a = 3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·3·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*3}=\frac{0}{6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*3}=\frac{8}{6}
=1+1/3
$