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m^2=16-16m
We move all terms to the left:
m^2-(16-16m)=0
We add all the numbers together, and all the variables
m^2-(-16m+16)=0
We get rid of parentheses
m^2+16m-16=0
a = 1; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·1·(-16)
Δ = 320
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{5}}{2*1}=\frac{-16-8\sqrt{5}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{5}}{2*1}=\frac{-16+8\sqrt{5}}{2} $
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