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

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

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

We multiply parentheses ..

(+12t^2-4t+3t-1)=0

We get rid of parentheses

12t^2-4t+3t-1=0

We add all the numbers together, and all the variables

12t^2-1t-1=0

a = 12; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·12·(-1)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*12}=\frac{-6}{24}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*12}=\frac{8}{24}
=1/3
$