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m(2)=12
We move all terms to the left:
m(2)-(12)=0
We add all the numbers together, and all the variables
m^2-12=0
a = 1; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·1·(-12)
Δ = 48
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*1}=\frac{0-4\sqrt{3}}{2} =-\frac{4\sqrt{3}}{2} =-2\sqrt{3} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*1}=\frac{0+4\sqrt{3}}{2} =\frac{4\sqrt{3}}{2} =2\sqrt{3} $
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