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1m^2+2m=4
We move all terms to the left:
1m^2+2m-(4)=0
We add all the numbers together, and all the variables
m^2+2m-4=0
a = 1; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·1·(-4)
Δ = 20
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*1}=\frac{-2-2\sqrt{5}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*1}=\frac{-2+2\sqrt{5}}{2} $
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