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16z^2+33z+2=0
a = 16; b = 33; c = +2;
Δ = b2-4ac
Δ = 332-4·16·2
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-31}{2*16}=\frac{-64}{32} =-2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+31}{2*16}=\frac{-2}{32} =-1/16 $
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