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m^2+6m-2=0
a = 1; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·1·(-2)
Δ = 44
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{11}}{2*1}=\frac{-6-2\sqrt{11}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{11}}{2*1}=\frac{-6+2\sqrt{11}}{2} $
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