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4c-c^2=0
We add all the numbers together, and all the variables
-1c^2+4c=0
a = -1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1)·0
Δ = 16
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{16}=4$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-1}=\frac{-8}{-2} =+4 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1}=\frac{0}{-2} =0 $
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