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5z^2-25z+30=0
a = 5; b = -25; c = +30;
Δ = b2-4ac
Δ = -252-4·5·30
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$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5}{2*5}=\frac{20}{10} =2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5}{2*5}=\frac{30}{10} =3 $
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