3k^2-62+3k=0

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Solution for 3k^2-62+3k=0 equation: 



3k^2-62+3k=0

a = 3; b = 3; c = -62;
Δ = b2-4ac
Δ = 32-4·3·(-62)
Δ = 753
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{753}}{2*3}=\frac{-3-\sqrt{753}}{6}
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{753}}{2*3}=\frac{-3+\sqrt{753}}{6}
$

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