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5k^2-16k+12=0
a = 5; b = -16; c = +12;
Δ = b2-4ac
Δ = -162-4·5·12
Δ = 16
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{16}=4$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*5}=\frac{12}{10} =1+1/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*5}=\frac{20}{10} =2 $
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