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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4w^2+w-3=0

a = 4; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·4·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*4}=\frac{-8}{8}
=-1
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*4}=\frac{6}{8}
=3/4
$