(6w^2+7w)/(-4w^2-2w+3)=0

Solution for (6w^2+7w)/(-4w^2-2w+3)=0 equation:

(6w^2+7w)/(-4w^2-2w+3)=0


Domain of the equation: (-4w^2-2w+3)!=0

We move all terms containing w to the left, all other terms to the right

-4w^2-2w!=-3

w∈R


We multiply all the terms by the denominator


(6w^2+7w)=0

We get rid of parentheses

6w^2+7w=0

a = 6; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·6·0
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{-(7)-7}{2*6}=\frac{-14}{12}
=-1+1/6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*6}=\frac{0}{12}
=0
$