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16y^2-49=0
a = 16; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·16·(-49)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3136}=56$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-56}{2*16}=\frac{-56}{32} =-1+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+56}{2*16}=\frac{56}{32} =1+3/4 $
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