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b^2-400=1200
We move all terms to the left:
b^2-400-(1200)=0
We add all the numbers together, and all the variables
b^2-1600=0
a = 1; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·1·(-1600)
Δ = 6400
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{6400}=80$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80}{2*1}=\frac{-80}{2} =-40 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80}{2*1}=\frac{80}{2} =40 $
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