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25c+10c^2=0.
a = 10; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·10·0
Δ = 625
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{625}=25$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*10}=\frac{-50}{20} =-2+1/2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*10}=\frac{0}{20} =0 $
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