(5t-3)(t+12)=0

Solution for (5t-3)(t+12)=0 equation:

(5t-3)(t+12)=0

We multiply parentheses ..

(+5t^2+60t-3t-36)=0

We get rid of parentheses

5t^2+60t-3t-36=0

We add all the numbers together, and all the variables

5t^2+57t-36=0

a = 5; b = 57; c = -36;
Δ = b2-4ac
Δ = 572-4·5·(-36)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-63}{2*5}=\frac{-120}{10}
=-12
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+63}{2*5}=\frac{6}{10}
=3/5
$