-6w^2+11w-3=0

Solution for -6w^2+11w-3=0 equation:

-6w^2+11w-3=0

a = -6; b = 11; c = -3;
Δ = b2-4ac
Δ = 112-4·(-6)·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*-6}=\frac{-18}{-12}
=1+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*-6}=\frac{-4}{-12}
=1/3
$