(4c-3)(6c+1)=0

Solution for (4c-3)(6c+1)=0 equation:

(4c-3)(6c+1)=0

We multiply parentheses ..

(+24c^2+4c-18c-3)=0

We get rid of parentheses

24c^2+4c-18c-3=0

We add all the numbers together, and all the variables

24c^2-14c-3=0

a = 24; b = -14; c = -3;
Δ = b2-4ac
Δ = -142-4·24·(-3)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-22}{2*24}=\frac{-8}{48}
=-1/6
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+22}{2*24}=\frac{36}{48}
=3/4
$