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36p-4p^2=0
a = -4; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·(-4)·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*-4}=\frac{-72}{-8} =+9 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*-4}=\frac{0}{-8} =0 $
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