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4z^2+18z-52=0
a = 4; b = 18; c = -52;
Δ = b2-4ac
Δ = 182-4·4·(-52)
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-34}{2*4}=\frac{-52}{8} =-6+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+34}{2*4}=\frac{16}{8} =2 $
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