6c^2+10c=53

Solution for 6c^2+10c=53 equation:

6c^2+10c=53

We move all terms to the left:

6c^2+10c-(53)=0

a = 6; b = 10; c = -53;
Δ = b2-4ac
Δ = 102-4·6·(-53)
Δ = 1372
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:

$\sqrt{\Delta}=\sqrt{1372}=\sqrt{196*7}=\sqrt{196}*\sqrt{7}=14\sqrt{7}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14\sqrt{7}}{2*6}=\frac{-10-14\sqrt{7}}{12}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14\sqrt{7}}{2*6}=\frac{-10+14\sqrt{7}}{12}
$