-0.02x^2+3.6x+196=0

Solution for -0.02x^2+3.6x+196=0 equation:

-0.02x^2+3.6x+196=0

a = -0.02; b = 3.6; c = +196;
Δ = b2-4ac
Δ = 3.62-4·(-0.02)·196
Δ = 28.64
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{-(3.6)-\sqrt{28.64}}{2*-0.02}=\frac{-3.6-\sqrt{28.64}}{-0.04}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.6)+\sqrt{28.64}}{2*-0.02}=\frac{-3.6+\sqrt{28.64}}{-0.04}
$