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0.25+k+k^2=1
We move all terms to the left:
0.25+k+k^2-(1)=0
We add all the numbers together, and all the variables
k^2+k-0.75=0
a = 1; b = 1; c = -0.75;
Δ = b2-4ac
Δ = 12-4·1·(-0.75)
Δ = 4
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{4}=2$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-2}{2*1}=\frac{-3}{2} =-1+1/2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+2}{2*1}=\frac{1}{2} =1/2 $
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