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3m^2+7=136
We move all terms to the left:
3m^2+7-(136)=0
We add all the numbers together, and all the variables
3m^2-129=0
a = 3; b = 0; c = -129;
Δ = b2-4ac
Δ = 02-4·3·(-129)
Δ = 1548
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1548}=\sqrt{36*43}=\sqrt{36}*\sqrt{43}=6\sqrt{43}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{43}}{2*3}=\frac{0-6\sqrt{43}}{6} =-\frac{6\sqrt{43}}{6} =-\sqrt{43} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{43}}{2*3}=\frac{0+6\sqrt{43}}{6} =\frac{6\sqrt{43}}{6} =\sqrt{43} $
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