-5t2+45t+50=0

Solution for -5t2+45t+50=0 equation:

-5t^2+45t+50=0

a = -5; b = 45; c = +50;
Δ = b2-4ac
Δ = 452-4·(-5)·50
Δ = 3025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3025}=55$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-55}{2*-5}=\frac{-100}{-10}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+55}{2*-5}=\frac{10}{-10}
=-1
$