7t^2+20t-140=0

Solution for 7t^2+20t-140=0 equation:

7t^2+20t-140=0

a = 7; b = 20; c = -140;
Δ = b2-4ac
Δ = 202-4·7·(-140)
Δ = 4320
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:

$\sqrt{\Delta}=\sqrt{4320}=\sqrt{144*30}=\sqrt{144}*\sqrt{30}=12\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{30}}{2*7}=\frac{-20-12\sqrt{30}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{30}}{2*7}=\frac{-20+12\sqrt{30}}{14}
$