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6m^2=20
We move all terms to the left:
6m^2-(20)=0
a = 6; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·6·(-20)
Δ = 480
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*6}=\frac{0-4\sqrt{30}}{12} =-\frac{4\sqrt{30}}{12} =-\frac{\sqrt{30}}{3} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*6}=\frac{0+4\sqrt{30}}{12} =\frac{4\sqrt{30}}{12} =\frac{\sqrt{30}}{3} $
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