4=-16t^2+18t

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

4=-16t^2+18t

We move all terms to the left:

4-(-16t^2+18t)=0

We get rid of parentheses

16t^2-18t+4=0

a = 16; b = -18; c = +4;
Δ = b2-4ac
Δ = -182-4·16·4
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{17}}{2*16}=\frac{18-2\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{17}}{2*16}=\frac{18+2\sqrt{17}}{32}
$