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100-4p^2=0
a = -4; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-4)·100
Δ = 1600
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{1600}=40$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-4}=\frac{-40}{-8} =+5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-4}=\frac{40}{-8} =-5 $
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