5c^2-16c+12=0

Solution for 5c^2-16c+12=0 equation:

5c^2-16c+12=0

a = 5; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·5·12
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*5}=\frac{12}{10}
=1+1/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*5}=\frac{20}{10}
=2
$