.03y^2+.08y+.02=0

Solution for .03y^2+.08y+.02=0 equation:

.03y^2+.08y+.02=0

We add all the numbers together, and all the variables

.03y^2+.08y+0.02=0

a = .03; b = .08; c = +0.02;
Δ = b2-4ac
Δ = .082-4·.03·0.02
Δ = 0.004
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.08)-\sqrt{0.004}}{2*.03}=\frac{-0.08-\sqrt{0.004}}{0.06}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.08)+\sqrt{0.004}}{2*.03}=\frac{-0.08+\sqrt{0.004}}{0.06}
$