-4w^2-4w+3=0

Solution for -4w^2-4w+3=0 equation:

-4w^2-4w+3=0

a = -4; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·(-4)·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-4}=\frac{-4}{-8}
=1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-4}=\frac{12}{-8}
=-1+1/2
$