10c2-23c+12=0

Solution for 10c2-23c+12=0 equation:

10c^2-23c+12=0

a = 10; b = -23; c = +12;
Δ = b2-4ac
Δ = -232-4·10·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*10}=\frac{16}{20}
=4/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*10}=\frac{30}{20}
=1+1/2
$