X^2+0.042x-0.00012=0

Solution for X^2+0.042x-0.00012=0 equation:

X^2+0.042X-0.00012=0

a = 1; b = 0.042; c = -0.00012;
Δ = b2-4ac
Δ = 0.0422-4·1·(-0.00012)
Δ = 0.002244
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{0.002244}}{2*1}=\frac{-0.042-\sqrt{0.002244}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.042)+\sqrt{0.002244}}{2*1}=\frac{-0.042+\sqrt{0.002244}}{2}
$