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2k^2+14k+48=k^2+8
We move all terms to the left:
2k^2+14k+48-(k^2+8)=0
We get rid of parentheses
2k^2-k^2+14k-8+48=0
We add all the numbers together, and all the variables
k^2+14k+40=0
a = 1; b = 14; c = +40;
Δ = b2-4ac
Δ = 142-4·1·40
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6}{2*1}=\frac{-20}{2} =-10 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6}{2*1}=\frac{-8}{2} =-4 $
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