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n(n-3)=154
We move all terms to the left:
n(n-3)-(154)=0
We multiply parentheses
n^2-3n-154=0
a = 1; b = -3; c = -154;
Δ = b2-4ac
Δ = -32-4·1·(-154)
Δ = 625
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{625}=25$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-25}{2*1}=\frac{-22}{2} =-11 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+25}{2*1}=\frac{28}{2} =14 $
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