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2p^2-50=0
a = 2; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·2·(-50)
Δ = 400
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{400}=20$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*2}=\frac{-20}{4} =-5 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*2}=\frac{20}{4} =5 $
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