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4k^2-12k+8=0
a = 4; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·4·8
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*4}=\frac{8}{8} =1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*4}=\frac{16}{8} =2 $
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