5k^2+48k-20=0

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Solution for 5k^2+48k-20=0 equation:



5k^2+48k-20=0
a = 5; b = 48; c = -20;
Δ = b2-4ac
Δ = 482-4·5·(-20)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-52}{2*5}=\frac{-100}{10} =-10 $
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+52}{2*5}=\frac{4}{10} =2/5 $

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