(3t+1)(t-4)=0

Solution for (3t+1)(t-4)=0 equation:

(3t+1)(t-4)=0

We multiply parentheses ..

(+3t^2-12t+t-4)=0

We get rid of parentheses

3t^2-12t+t-4=0

We add all the numbers together, and all the variables

3t^2-11t-4=0

a = 3; b = -11; c = -4;
Δ = b2-4ac
Δ = -112-4·3·(-4)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*3}=\frac{-2}{6}
=-1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*3}=\frac{24}{6}
=4
$