3=4w^2+w

Solution for 3=4w^2+w equation:

3=4w^2+w

We move all terms to the left:

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

We get rid of parentheses

-4w^2-w+3=0

We add all the numbers together, and all the variables

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