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16^2=3.14r^2
We move all terms to the left:
16^2-(3.14r^2)=0
We add all the numbers together, and all the variables
-(3.14r^2)+256=0
We get rid of parentheses
-3.14r^2+256=0
a = -3.14; b = 0; c = +256;
Δ = b2-4ac
Δ = 02-4·(-3.14)·256
Δ = 3215.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{3215.36}}{2*-3.14}=\frac{0-\sqrt{3215.36}}{-6.28} =-\frac{\sqrt{}}{-6.28} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{3215.36}}{2*-3.14}=\frac{0+\sqrt{3215.36}}{-6.28} =\frac{\sqrt{}}{-6.28} $
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