-4(2w-5);w=-3

Solution for -4(2w-5);w=-3 equation:

-4(2w-5)w=-3

We move all terms to the left:

-4(2w-5)w-(-3)=0

We add all the numbers together, and all the variables

-4(2w-5)w+3=0

We multiply parentheses

-8w^2+20w+3=0

a = -8; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-8)·3
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{31}}{2*-8}=\frac{-20-4\sqrt{31}}{-16}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{31}}{2*-8}=\frac{-20+4\sqrt{31}}{-16}
$