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2z^2+10z+8=0
a = 2; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·2·8
Δ = 36
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{36}=6$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*2}=\frac{-16}{4} =-4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*2}=\frac{-4}{4} =-1 $
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