c(4c+15)+4=0

Solution for c(4c+15)+4=0 equation:

c(4c+15)+4=0

We multiply parentheses

4c^2+15c+4=0

a = 4; b = 15; c = +4;
Δ = b2-4ac
Δ = 152-4·4·4
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{161}}{2*4}=\frac{-15-\sqrt{161}}{8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{161}}{2*4}=\frac{-15+\sqrt{161}}{8}
$