-4w^2+33w-35=0

Solution for -4w^2+33w-35=0 equation:

-4w^2+33w-35=0

a = -4; b = 33; c = -35;
Δ = b2-4ac
Δ = 332-4·(-4)·(-35)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-23}{2*-4}=\frac{-56}{-8}
=+7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+23}{2*-4}=\frac{-10}{-8}
=1+1/4
$