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400=m^2
We move all terms to the left:
400-(m^2)=0
We add all the numbers together, and all the variables
-1m^2+400=0
a = -1; b = 0; c = +400;
Δ = b2-4ac
Δ = 02-4·(-1)·400
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1600}=40$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-1}=\frac{-40}{-2} =+20 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-1}=\frac{40}{-2} =-20 $
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