t2-14t-98=0

Solution for t2-14t-98=0 equation:

t2-14t-98=0

We add all the numbers together, and all the variables

t^2-14t-98=0

a = 1; b = -14; c = -98;
Δ = b2-4ac
Δ = -142-4·1·(-98)
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14\sqrt{3}}{2*1}=\frac{14-14\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14\sqrt{3}}{2*1}=\frac{14+14\sqrt{3}}{2}
$