0.1x^2+1.2x+3.5=0

Solution for 0.1x^2+1.2x+3.5=0 equation:

0.1x^2+1.2x+3.5=0

a = 0.1; b = 1.2; c = +3.5;
Δ = b2-4ac
Δ = 1.22-4·0.1·3.5
Δ = 0.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.2)-\sqrt{0.04}}{2*0.1}=\frac{-1.2-\sqrt{0.04}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+\sqrt{0.04}}{2*0.1}=\frac{-1.2+\sqrt{0.04}}{0.2}
$