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2z^2+13z+18=0
a = 2; b = 13; c = +18;
Δ = b2-4ac
Δ = 132-4·2·18
Δ = 25
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{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*2}=\frac{-18}{4} =-4+1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*2}=\frac{-8}{4} =-2 $
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