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3m^2+6=10
We move all terms to the left:
3m^2+6-(10)=0
We add all the numbers together, and all the variables
3m^2-4=0
a = 3; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·3·(-4)
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*3}=\frac{0-4\sqrt{3}}{6} =-\frac{4\sqrt{3}}{6} =-\frac{2\sqrt{3}}{3} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*3}=\frac{0+4\sqrt{3}}{6} =\frac{4\sqrt{3}}{6} =\frac{2\sqrt{3}}{3} $
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