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3600+b2=10000
We move all terms to the left:
3600+b2-(10000)=0
We add all the numbers together, and all the variables
b^2-6400=0
a = 1; b = 0; c = -6400;
Δ = b2-4ac
Δ = 02-4·1·(-6400)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25600}=160$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160}{2*1}=\frac{-160}{2} =-80 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160}{2*1}=\frac{160}{2} =80 $
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