2x*3-3x^2+1=0

Solution for 2x*3-3x^2+1=0 equation:

2x*3-3x^2+1=0

Wy multiply elements

-3x^2+6x+1=0

a = -3; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-3)·1
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*-3}=\frac{-6-4\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*-3}=\frac{-6+4\sqrt{3}}{-6}
$