-25=-0.03*x^2+x

Solution for -25=-0.03*x^2+x equation:

-25=-0.03x^2+x

We move all terms to the left:

-25-(-0.03x^2+x)=0

We get rid of parentheses

0.03x^2-x-25=0

We add all the numbers together, and all the variables

0.03x^2-1x-25=0

a = 0.03; b = -1; c = -25;
Δ = b2-4ac
Δ = -12-4·0.03·(-25)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-2}{2*0.03}=\frac{-1}{0.06}
=-16+0.04/0.06
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+2}{2*0.03}=\frac{3}{0.06}
=50
$