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10k^2-12k-10=0
a = 10; b = -12; c = -10;
Δ = b2-4ac
Δ = -122-4·10·(-10)
Δ = 544
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{34}}{2*10}=\frac{12-4\sqrt{34}}{20} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{34}}{2*10}=\frac{12+4\sqrt{34}}{20} $
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