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4z^2+11z=0
a = 4; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·4·0
Δ = 121
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{121}=11$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*4}=\frac{-22}{8} =-2+3/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*4}=\frac{0}{8} =0 $
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