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k^2+4k-45=2
We move all terms to the left:
k^2+4k-45-(2)=0
We add all the numbers together, and all the variables
k^2+4k-47=0
a = 1; b = 4; c = -47;
Δ = b2-4ac
Δ = 42-4·1·(-47)
Δ = 204
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{51}}{2*1}=\frac{-4-2\sqrt{51}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{51}}{2*1}=\frac{-4+2\sqrt{51}}{2} $
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