15t(t+1)=t-3

Solution for 15t(t+1)=t-3 equation:

15t(t+1)=t-3

We move all terms to the left:

15t(t+1)-(t-3)=0

We multiply parentheses

15t^2+15t-(t-3)=0

We get rid of parentheses

15t^2+15t-t+3=0

We add all the numbers together, and all the variables

15t^2+14t+3=0

a = 15; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·15·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*15}=\frac{-18}{30}
=-3/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*15}=\frac{-10}{30}
=-1/3
$