15+10t-5t^2=0

Solution for 15+10t-5t^2=0 equation:

15+10t-5t^2=0

a = -5; b = 10; c = +15;
Δ = b2-4ac
Δ = 102-4·(-5)·15
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-20}{2*-5}=\frac{-30}{-10}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+20}{2*-5}=\frac{10}{-10}
=-1
$