5+34t-16t^2=0

Solution for 5+34t-16t^2=0 equation:

5+34t-16t^2=0

a = -16; b = 34; c = +5;
Δ = b2-4ac
Δ = 342-4·(-16)·5
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-6\sqrt{41}}{2*-16}=\frac{-34-6\sqrt{41}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+6\sqrt{41}}{2*-16}=\frac{-34+6\sqrt{41}}{-32}
$