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4c(c-4)=9c+5
We move all terms to the left:
4c(c-4)-(9c+5)=0
We multiply parentheses
4c^2-16c-(9c+5)=0
We get rid of parentheses
4c^2-16c-9c-5=0
We add all the numbers together, and all the variables
4c^2-25c-5=0
a = 4; b = -25; c = -5;
Δ = b2-4ac
Δ = -252-4·4·(-5)
Δ = 705
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{-(-25)-\sqrt{705}}{2*4}=\frac{25-\sqrt{705}}{8} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{705}}{2*4}=\frac{25+\sqrt{705}}{8} $
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