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c^2=2c^2-12c+20
We move all terms to the left:
c^2-(2c^2-12c+20)=0
We get rid of parentheses
c^2-2c^2+12c-20=0
We add all the numbers together, and all the variables
-1c^2+12c-20=0
a = -1; b = 12; c = -20;
Δ = b2-4ac
Δ = 122-4·(-1)·(-20)
Δ = 64
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*-1}=\frac{-20}{-2} =+10 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*-1}=\frac{-4}{-2} =+2 $
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