5k2+21k+18=0

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Solution for 5k2+21k+18=0 equation:



5k^2+21k+18=0
a = 5; b = 21; c = +18;
Δ = b2-4ac
Δ = 212-4·5·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-9}{2*5}=\frac{-30}{10} =-3 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+9}{2*5}=\frac{-12}{10} =-1+1/5 $

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