3c^2-32c+20=0

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Solution for 3c^2-32c+20=0 equation:



3c^2-32c+20=0
a = 3; b = -32; c = +20;
Δ = b2-4ac
Δ = -322-4·3·20
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-28}{2*3}=\frac{4}{6} =2/3 $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+28}{2*3}=\frac{60}{6} =10 $

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