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5(c-6)c=c+11(24)
We move all terms to the left:
5(c-6)c-(c+11(24))=0
We multiply parentheses
5c^2-30c-(c+1124)=0
We get rid of parentheses
5c^2-30c-c-1124=0
We add all the numbers together, and all the variables
5c^2-31c-1124=0
a = 5; b = -31; c = -1124;
Δ = b2-4ac
Δ = -312-4·5·(-1124)
Δ = 23441
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{-(-31)-\sqrt{23441}}{2*5}=\frac{31-\sqrt{23441}}{10} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{23441}}{2*5}=\frac{31+\sqrt{23441}}{10} $
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