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2c-(c-5)=c(2c-5)-5
We move all terms to the left:
2c-(c-5)-(c(2c-5)-5)=0
We get rid of parentheses
2c-c-(c(2c-5)-5)+5=0
We calculate terms in parentheses: -(c(2c-5)-5), so:We add all the numbers together, and all the variables
c(2c-5)-5
We multiply parentheses
2c^2-5c-5
Back to the equation:
-(2c^2-5c-5)
c-(2c^2-5c-5)+5=0
We get rid of parentheses
-2c^2+c+5c+5+5=0
We add all the numbers together, and all the variables
-2c^2+6c+10=0
a = -2; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-2)·10
Δ = 116
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{29}}{2*-2}=\frac{-6-2\sqrt{29}}{-4} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{29}}{2*-2}=\frac{-6+2\sqrt{29}}{-4} $
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