15w^2+28w+3=-2

Solution for 15w^2+28w+3=-2 equation:

15w^2+28w+3=-2

We move all terms to the left:

15w^2+28w+3-(-2)=0

We add all the numbers together, and all the variables

15w^2+28w+5=0

a = 15; b = 28; c = +5;
Δ = b2-4ac
Δ = 282-4·15·5
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-22}{2*15}=\frac{-50}{30}
=-1+2/3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+22}{2*15}=\frac{-6}{30}
=-1/5
$