0.012x^2+0.1x+0.16=0

Solution for 0.012x^2+0.1x+0.16=0 equation:

0.012x^2+0.1x+0.16=0

a = 0.012; b = 0.1; c = +0.16;
Δ = b2-4ac
Δ = 0.12-4·0.012·0.16
Δ = 0.00232
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{-(0.1)-\sqrt{0.00232}}{2*0.012}=\frac{-0.1-\sqrt{0.00232}}{0.024}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{0.00232}}{2*0.012}=\frac{-0.1+\sqrt{0.00232}}{0.024}
$