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16r^2=400
We move all terms to the left:
16r^2-(400)=0
a = 16; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·16·(-400)
Δ = 25600
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{25600}=160$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160}{2*16}=\frac{-160}{32} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160}{2*16}=\frac{160}{32} =5 $
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