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16u^2-9=0
a = 16; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·16·(-9)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*16}=\frac{-24}{32} =-3/4 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*16}=\frac{24}{32} =3/4 $
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