3x^2+0.042x-0.000191=0.015

Solution for 3x^2+0.042x-0.000191=0.015 equation:

3x^2+0.042x-0.000191=0.015

We move all terms to the left:

3x^2+0.042x-0.000191-(0.015)=0

We add all the numbers together, and all the variables

3x^2+0.042x-0.015191=0

a = 3; b = 0.042; c = -0.015191;
Δ = b2-4ac
Δ = 0.0422-4·3·(-0.015191)
Δ = 0.184056
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{-(0.042)-\sqrt{0.184056}}{2*3}=\frac{-0.042-\sqrt{0.184056}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.042)+\sqrt{0.184056}}{2*3}=\frac{-0.042+\sqrt{0.184056}}{6}
$