4t^2+20t-84=0

Solution for 4t^2+20t-84=0 equation:

4t^2+20t-84=0

a = 4; b = 20; c = -84;
Δ = b2-4ac
Δ = 202-4·4·(-84)
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{109}}{2*4}=\frac{-20-4\sqrt{109}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{109}}{2*4}=\frac{-20+4\sqrt{109}}{8}
$