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5k^2-27k+10=0
a = 5; b = -27; c = +10;
Δ = b2-4ac
Δ = -272-4·5·10
Δ = 529
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{529}=23$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-23}{2*5}=\frac{4}{10} =2/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+23}{2*5}=\frac{50}{10} =5 $
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