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16v^2-25=0
a = 16; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·16·(-25)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*16}=\frac{-40}{32} =-1+1/4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*16}=\frac{40}{32} =1+1/4 $
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