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12m^2=7
We move all terms to the left:
12m^2-(7)=0
a = 12; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·12·(-7)
Δ = 336
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{21}}{2*12}=\frac{0-4\sqrt{21}}{24} =-\frac{4\sqrt{21}}{24} =-\frac{\sqrt{21}}{6} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{21}}{2*12}=\frac{0+4\sqrt{21}}{24} =\frac{4\sqrt{21}}{24} =\frac{\sqrt{21}}{6} $
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