2k2+22k+60=0

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Solution for 2k2+22k+60=0 equation: 



2k^2+22k+60=0

a = 2; b = 22; c = +60;
Δ = b2-4ac
Δ = 222-4·2·60
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*2}=\frac{-24}{4}
=-6
$


$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*2}=\frac{-20}{4}
=-5
$

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