33=8t^2+35t

Solution for 33=8t^2+35t equation:

33=8t^2+35t

We move all terms to the left:

33-(8t^2+35t)=0

We get rid of parentheses

-8t^2-35t+33=0

a = -8; b = -35; c = +33;
Δ = b2-4ac
Δ = -352-4·(-8)·33
Δ = 2281
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{-(-35)-\sqrt{2281}}{2*-8}=\frac{35-\sqrt{2281}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{2281}}{2*-8}=\frac{35+\sqrt{2281}}{-16}
$