c2+16c+60=0

Solution for c2+16c+60=0 equation:

c2+16c+60=0

We add all the numbers together, and all the variables

c^2+16c+60=0

a = 1; b = 16; c = +60;
Δ = b2-4ac
Δ = 162-4·1·60
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*1}=\frac{-20}{2}
=-10
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*1}=\frac{-12}{2}
=-6
$