80t+2-5t^2=0

Solution for 80t+2-5t^2=0 equation:

80t+2-5t^2=0

a = -5; b = 80; c = +2;
Δ = b2-4ac
Δ = 802-4·(-5)·2
Δ = 6440
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:

$\sqrt{\Delta}=\sqrt{6440}=\sqrt{4*1610}=\sqrt{4}*\sqrt{1610}=2\sqrt{1610}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-2\sqrt{1610}}{2*-5}=\frac{-80-2\sqrt{1610}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+2\sqrt{1610}}{2*-5}=\frac{-80+2\sqrt{1610}}{-10}
$