25c+25c^2-6=0

Solution for 25c+25c^2-6=0 equation:

25c+25c^2-6=0

a = 25; b = 25; c = -6;
Δ = b2-4ac
Δ = 252-4·25·(-6)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*25}=\frac{-60}{50}
=-1+1/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*25}=\frac{10}{50}
=1/5
$