-20x^2+3x+196=0

Solution for -20x^2+3x+196=0 equation:

-20x^2+3x+196=0

a = -20; b = 3; c = +196;
Δ = b2-4ac
Δ = 32-4·(-20)·196
Δ = 15689
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)-\sqrt{15689}}{2*-20}=\frac{-3-\sqrt{15689}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{15689}}{2*-20}=\frac{-3+\sqrt{15689}}{-40}
$