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n2+12n-133=0
We add all the numbers together, and all the variables
n^2+12n-133=0
a = 1; b = 12; c = -133;
Δ = b2-4ac
Δ = 122-4·1·(-133)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-26}{2*1}=\frac{-38}{2} =-19 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+26}{2*1}=\frac{14}{2} =7 $
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