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4p^2=100
We move all terms to the left:
4p^2-(100)=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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