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5k^2+10k-2=0
a = 5; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·5·(-2)
Δ = 140
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{35}}{2*5}=\frac{-10-2\sqrt{35}}{10} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{35}}{2*5}=\frac{-10+2\sqrt{35}}{10} $
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