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-c(c-3)=1
We move all terms to the left:
-c(c-3)-(1)=0
We multiply parentheses
-c^2+3c-1=0
We add all the numbers together, and all the variables
-1c^2+3c-1=0
a = -1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·(-1)·(-1)
Δ = 5
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{-(3)-\sqrt{5}}{2*-1}=\frac{-3-\sqrt{5}}{-2} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*-1}=\frac{-3+\sqrt{5}}{-2} $
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