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c^2=15c+44
We move all terms to the left:
c^2-(15c+44)=0
We get rid of parentheses
c^2-15c-44=0
a = 1; b = -15; c = -44;
Δ = b2-4ac
Δ = -152-4·1·(-44)
Δ = 401
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{401}}{2*1}=\frac{15-\sqrt{401}}{2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{401}}{2*1}=\frac{15+\sqrt{401}}{2} $
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