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24-(3c+4)=2(c+4)c
We move all terms to the left:
24-(3c+4)-(2(c+4)c)=0
We get rid of parentheses
-3c-(2(c+4)c)-4+24=0
We calculate terms in parentheses: -(2(c+4)c), so:We add all the numbers together, and all the variables
2(c+4)c
We multiply parentheses
2c^2+8c
Back to the equation:
-(2c^2+8c)
-3c-(2c^2+8c)+20=0
We get rid of parentheses
-2c^2-3c-8c+20=0
We add all the numbers together, and all the variables
-2c^2-11c+20=0
a = -2; b = -11; c = +20;
Δ = b2-4ac
Δ = -112-4·(-2)·20
Δ = 281
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{-(-11)-\sqrt{281}}{2*-2}=\frac{11-\sqrt{281}}{-4} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{281}}{2*-2}=\frac{11+\sqrt{281}}{-4} $
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