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11m^2+14=47
We move all terms to the left:
11m^2+14-(47)=0
We add all the numbers together, and all the variables
11m^2-33=0
a = 11; b = 0; c = -33;
Δ = b2-4ac
Δ = 02-4·11·(-33)
Δ = 1452
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1452}=\sqrt{484*3}=\sqrt{484}*\sqrt{3}=22\sqrt{3}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22\sqrt{3}}{2*11}=\frac{0-22\sqrt{3}}{22} =-\frac{22\sqrt{3}}{22} =-\sqrt{3} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22\sqrt{3}}{2*11}=\frac{0+22\sqrt{3}}{22} =\frac{22\sqrt{3}}{22} =\sqrt{3} $
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