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16=r2
We move all terms to the left:
16-(r2)=0
We add all the numbers together, and all the variables
-1r^2+16=0
a = -1; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-1)·16
Δ = 64
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-1}=\frac{-8}{-2} =+4 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-1}=\frac{8}{-2} =-4 $
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