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25b^2-16=0
a = 25; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·25·(-16)
Δ = 1600
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{1600}=40$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*25}=\frac{-40}{50} =-4/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*25}=\frac{40}{50} =4/5 $
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