t^2+40t-196=0

Solution for t^2+40t-196=0 equation:

t^2+40t-196=0

a = 1; b = 40; c = -196;
Δ = b2-4ac
Δ = 402-4·1·(-196)
Δ = 2384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2384}=\sqrt{16*149}=\sqrt{16}*\sqrt{149}=4\sqrt{149}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{149}}{2*1}=\frac{-40-4\sqrt{149}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{149}}{2*1}=\frac{-40+4\sqrt{149}}{2}
$