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3=k2
We move all terms to the left:
3-(k2)=0
We add all the numbers together, and all the variables
-1k^2+3=0
a = -1; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-1)·3
Δ = 12
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{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-1}=\frac{0-2\sqrt{3}}{-2} =-\frac{2\sqrt{3}}{-2} =-\frac{\sqrt{3}}{-1}k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-1}=\frac{0+2\sqrt{3}}{-2} =\frac{2\sqrt{3}}{-2} =\frac{\sqrt{3}}{-1}
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