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c^2+18c+65=9
We move all terms to the left:
c^2+18c+65-(9)=0
We add all the numbers together, and all the variables
c^2+18c+56=0
a = 1; b = 18; c = +56;
Δ = b2-4ac
Δ = 182-4·1·56
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-10}{2*1}=\frac{-28}{2} =-14 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+10}{2*1}=\frac{-8}{2} =-4 $
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