13t^2+20t-188=0

Solution for 13t^2+20t-188=0 equation:

13t^2+20t-188=0

a = 13; b = 20; c = -188;
Δ = b2-4ac
Δ = 202-4·13·(-188)
Δ = 10176
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:

$\sqrt{\Delta}=\sqrt{10176}=\sqrt{64*159}=\sqrt{64}*\sqrt{159}=8\sqrt{159}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{159}}{2*13}=\frac{-20-8\sqrt{159}}{26}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{159}}{2*13}=\frac{-20+8\sqrt{159}}{26}
$