(4w+1)(3-w)=0

Solution for (4w+1)(3-w)=0 equation:

(4w+1)(3-w)=0

We add all the numbers together, and all the variables

(4w+1)(-1w+3)=0

We multiply parentheses ..

(-4w^2+12w-1w+3)=0

We get rid of parentheses

-4w^2+12w-1w+3=0

We add all the numbers together, and all the variables

-4w^2+11w+3=0

a = -4; b = 11; c = +3;
Δ = b2-4ac
Δ = 112-4·(-4)·3
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*-4}=\frac{-24}{-8}
=+3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*-4}=\frac{2}{-8}
=-1/4
$