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5c^2+20c^2-160c=0
We add all the numbers together, and all the variables
25c^2-160c=0
a = 25; b = -160; c = 0;
Δ = b2-4ac
Δ = -1602-4·25·0
Δ = 25600
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{25600}=160$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-160}{2*25}=\frac{0}{50} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+160}{2*25}=\frac{320}{50} =6+2/5 $
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