-5t^2+30t+35=0

Solution for -5t^2+30t+35=0 equation:

-5t^2+30t+35=0

a = -5; b = 30; c = +35;
Δ = b2-4ac
Δ = 302-4·(-5)·35
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-40}{2*-5}=\frac{-70}{-10}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+40}{2*-5}=\frac{10}{-10}
=-1
$