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5z^2+26z+5=0
a = 5; b = 26; c = +5;
Δ = b2-4ac
Δ = 262-4·5·5
Δ = 576
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{576}=24$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-24}{2*5}=\frac{-50}{10} =-5 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+24}{2*5}=\frac{-2}{10} =-1/5 $
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