5t^2+20t+15=0

Solution for 5t^2+20t+15=0 equation:

5t^2+20t+15=0

a = 5; b = 20; c = +15;
Δ = b2-4ac
Δ = 202-4·5·15
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*5}=\frac{-30}{10}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*5}=\frac{-10}{10}
=-1
$