(3c+4)(7c-6)=0

Solution for (3c+4)(7c-6)=0 equation:

(3c+4)(7c-6)=0

We multiply parentheses ..

(+21c^2-18c+28c-24)=0

We get rid of parentheses

21c^2-18c+28c-24=0

We add all the numbers together, and all the variables

21c^2+10c-24=0

a = 21; b = 10; c = -24;
Δ = b2-4ac
Δ = 102-4·21·(-24)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-46}{2*21}=\frac{-56}{42}
=-1+1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+46}{2*21}=\frac{36}{42}
=6/7
$