-14w^2+20w-3=0

Solution for -14w^2+20w-3=0 equation:

-14w^2+20w-3=0

a = -14; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·(-14)·(-3)
Δ = 232
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{58}}{2*-14}=\frac{-20-2\sqrt{58}}{-28}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{58}}{2*-14}=\frac{-20+2\sqrt{58}}{-28}
$