5t^2+20t-585=0

Solution for 5t^2+20t-585=0 equation:

5t^2+20t-585=0

a = 5; b = 20; c = -585;
Δ = b2-4ac
Δ = 202-4·5·(-585)
Δ = 12100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12100}=110$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-110}{2*5}=\frac{-130}{10}
=-13
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+110}{2*5}=\frac{90}{10}
=9
$