t2+3t-98=0

Solution for t2+3t-98=0 equation:

t2+3t-98=0

We add all the numbers together, and all the variables

t^2+3t-98=0

a = 1; b = 3; c = -98;
Δ = b2-4ac
Δ = 32-4·1·(-98)
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{401}}{2*1}=\frac{-3-\sqrt{401}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{401}}{2*1}=\frac{-3+\sqrt{401}}{2}
$