0.2x^2+9x+1300=3300

Solution for 0.2x^2+9x+1300=3300 equation:

0.2x^2+9x+1300=3300

We move all terms to the left:

0.2x^2+9x+1300-(3300)=0

We add all the numbers together, and all the variables

0.2x^2+9x-2000=0

a = 0.2; b = 9; c = -2000;
Δ = b2-4ac
Δ = 92-4·0.2·(-2000)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-41}{2*0.2}=\frac{-50}{0.4}
=-125
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+41}{2*0.2}=\frac{32}{0.4}
=80
$