10c^2-19c+6=0

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Solution for 10c^2-19c+6=0 equation: 



10c^2-19c+6=0

a = 10; b = -19; c = +6;
Δ = b2-4ac
Δ = -192-4·10·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$


$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*10}=\frac{8}{20}
=2/5
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*10}=\frac{30}{20}
=1+1/2
$

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