0=0.0012x^2-0.50x+25.98

Solution for 0=0.0012x^2-0.50x+25.98 equation:

0=0.0012x^2-0.50x+25.98

We move all terms to the left:

0-(0.0012x^2-0.50x+25.98)=0

We add all the numbers together, and all the variables

-(0.0012x^2-0.50x+25.98)=0

We get rid of parentheses

-0.0012x^2+0.50x-25.98=0

a = -0.0012; b = 0.50; c = -25.98;
Δ = b2-4ac
Δ = 0.502-4·(-0.0012)·(-25.98)
Δ = 0.125296
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{-(0.50)-\sqrt{0.125296}}{2*-0.0012}=\frac{-0.5-\sqrt{0.125296}}{-0.0024}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.50)+\sqrt{0.125296}}{2*-0.0012}=\frac{-0.5+\sqrt{0.125296}}{-0.0024}
$