6+45t-16t^2=0

Solution for 6+45t-16t^2=0 equation:

6+45t-16t^2=0

a = -16; b = 45; c = +6;
Δ = b2-4ac
Δ = 452-4·(-16)·6
Δ = 2409
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{-(45)-\sqrt{2409}}{2*-16}=\frac{-45-\sqrt{2409}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+\sqrt{2409}}{2*-16}=\frac{-45+\sqrt{2409}}{-32}
$