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x2+2x+5=125
We move all terms to the left:
x2+2x+5-(125)=0
We add all the numbers together, and all the variables
x^2+2x-120=0
a = 1; b = 2; c = -120;
Δ = b2-4ac
Δ = 22-4·1·(-120)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*1}=\frac{-24}{2} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*1}=\frac{20}{2} =10 $
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