130+12t-16t^2=0

Solution for 130+12t-16t^2=0 equation:

130+12t-16t^2=0

a = -16; b = 12; c = +130;
Δ = b2-4ac
Δ = 122-4·(-16)·130
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-92}{2*-16}=\frac{-104}{-32}
=3+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+92}{2*-16}=\frac{80}{-32}
=-2+1/2
$