(2c-7)(6c+1)=0

Solution for (2c-7)(6c+1)=0 equation:

(2c-7)(6c+1)=0

We multiply parentheses ..

(+12c^2+2c-42c-7)=0

We get rid of parentheses

12c^2+2c-42c-7=0

We add all the numbers together, and all the variables

12c^2-40c-7=0

a = 12; b = -40; c = -7;
Δ = b2-4ac
Δ = -402-4·12·(-7)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-44}{2*12}=\frac{-4}{24}
=-1/6
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+44}{2*12}=\frac{84}{24}
=3+1/2
$