(t+1)(-2t-3)=0

Solution for (t+1)(-2t-3)=0 equation:

(t+1)(-2t-3)=0

We multiply parentheses ..

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

We get rid of parentheses

-2t^2-3t-2t-3=0

We add all the numbers together, and all the variables

-2t^2-5t-3=0

a = -2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-2)·(-3)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-2}=\frac{4}{-4}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-2}=\frac{6}{-4}
=-1+1/2
$