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m^2+14m=49
We move all terms to the left:
m^2+14m-(49)=0
a = 1; b = 14; c = -49;
Δ = b2-4ac
Δ = 142-4·1·(-49)
Δ = 392
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14\sqrt{2}}{2*1}=\frac{-14-14\sqrt{2}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14\sqrt{2}}{2*1}=\frac{-14+14\sqrt{2}}{2} $
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