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25b^2+50b=0
a = 25; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·25·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2500}=50$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*25}=\frac{-100}{50} =-2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*25}=\frac{0}{50} =0 $
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