5t^2+14t-98=0

Solution for 5t^2+14t-98=0 equation:

5t^2+14t-98=0

a = 5; b = 14; c = -98;
Δ = b2-4ac
Δ = 142-4·5·(-98)
Δ = 2156
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:

$\sqrt{\Delta}=\sqrt{2156}=\sqrt{196*11}=\sqrt{196}*\sqrt{11}=14\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{11}}{2*5}=\frac{-14-14\sqrt{11}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{11}}{2*5}=\frac{-14+14\sqrt{11}}{10}
$