0,1y^2+y+2,4=0

Solution for 0,1y^2+y+2,4=0 equation:

0.1y^2+y+2.4=0

a = 0.1; b = 1; c = +2.4;
Δ = b2-4ac
Δ = 12-4·0.1·2.4
Δ = 0.04
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{-(1)-\sqrt{0.04}}{2*0.1}=\frac{-1-\sqrt{0.04}}{0.2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.04}}{2*0.1}=\frac{-1+\sqrt{0.04}}{0.2}
$