-42.5t^2+30t+6=0

Solution for -42.5t^2+30t+6=0 equation:

-42.5t^2+30t+6=0

a = -42.5; b = 30; c = +6;
Δ = b2-4ac
Δ = 302-4·(-42.5)·6
Δ = 1920
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:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-8\sqrt{30}}{2*-42.5}=\frac{-30-8\sqrt{30}}{-85}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+8\sqrt{30}}{2*-42.5}=\frac{-30+8\sqrt{30}}{-85}
$