4(-2t+5)(t+3)=0

Solution for 4(-2t+5)(t+3)=0 equation:

4(-2t+5)(t+3)=0

We multiply parentheses ..

4(-2t^2-6t+5t+15)=0

We multiply parentheses

-8t^2-24t+20t+60=0

We add all the numbers together, and all the variables

-8t^2-4t+60=0

a = -8; b = -4; c = +60;
Δ = b2-4ac
Δ = -42-4·(-8)·60
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-44}{2*-8}=\frac{-40}{-16}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+44}{2*-8}=\frac{48}{-16}
=-3
$