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a^2=0.16
We move all terms to the left:
a^2-(0.16)=0
We add all the numbers together, and all the variables
a^2-0.16=0
a = 1; b = 0; c = -0.16;
Δ = b2-4ac
Δ = 02-4·1·(-0.16)
Δ = 0.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.64}}{2*1}=\frac{0-\sqrt{0.64}}{2} =-\frac{\sqrt{}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.64}}{2*1}=\frac{0+\sqrt{0.64}}{2} =\frac{\sqrt{}}{2} $
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