3(4c^2-28c+49)=15

Solution for 3(4c^2-28c+49)=15 equation:

3(4c^2-28c+49)=15

We move all terms to the left:

3(4c^2-28c+49)-(15)=0

We multiply parentheses

12c^2-84c+147-15=0

We add all the numbers together, and all the variables

12c^2-84c+132=0

a = 12; b = -84; c = +132;
Δ = b2-4ac
Δ = -842-4·12·132
Δ = 720
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-12\sqrt{5}}{2*12}=\frac{84-12\sqrt{5}}{24}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+12\sqrt{5}}{2*12}=\frac{84+12\sqrt{5}}{24}
$