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p^2=0.36
We move all terms to the left:
p^2-(0.36)=0
We add all the numbers together, and all the variables
p^2-0.36=0
a = 1; b = 0; c = -0.36;
Δ = b2-4ac
Δ = 02-4·1·(-0.36)
Δ = 1.44
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{1.44}}{2*1}=\frac{0-\sqrt{1.44}}{2} =-\frac{\sqrt{}}{2} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.44}}{2*1}=\frac{0+\sqrt{1.44}}{2} =\frac{\sqrt{}}{2} $
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