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18z^2+49z+10=0
a = 18; b = 49; c = +10;
Δ = b2-4ac
Δ = 492-4·18·10
Δ = 1681
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{1681}=41$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-41}{2*18}=\frac{-90}{36} =-2+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+41}{2*18}=\frac{-8}{36} =-2/9 $
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