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16z^2-14z-15=0
a = 16; b = -14; c = -15;
Δ = b2-4ac
Δ = -142-4·16·(-15)
Δ = 1156
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{1156}=34$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-34}{2*16}=\frac{-20}{32} =-5/8 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+34}{2*16}=\frac{48}{32} =1+1/2 $
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