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1-25b^2=0
a = -25; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-25)·1
Δ = 100
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{100}=10$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-25}=\frac{-10}{-50} =1/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-25}=\frac{10}{-50} =-1/5 $
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