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(k+6)(k+4)=8
We move all terms to the left:
(k+6)(k+4)-(8)=0
We multiply parentheses ..
(+k^2+4k+6k+24)-8=0
We get rid of parentheses
k^2+4k+6k+24-8=0
We add all the numbers together, and all the variables
k^2+10k+16=0
a = 1; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·1·16
Δ = 36
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{36}=6$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*1}=\frac{-16}{2} =-8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*1}=\frac{-4}{2} =-2 $
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