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 = 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}
$