4+18t-16t^2=0

Solution for 4+18t-16t^2=0 equation:

4+18t-16t^2=0

a = -16; b = 18; c = +4;
Δ = b2-4ac
Δ = 182-4·(-16)·4
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{145}}{2*-16}=\frac{-18-2\sqrt{145}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{145}}{2*-16}=\frac{-18+2\sqrt{145}}{-32}
$