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n(n-3)=90
We move all terms to the left:
n(n-3)-(90)=0
We multiply parentheses
n^2-3n-90=0
a = 1; b = -3; c = -90;
Δ = b2-4ac
Δ = -32-4·1·(-90)
Δ = 369
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{41}}{2*1}=\frac{3-3\sqrt{41}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{41}}{2*1}=\frac{3+3\sqrt{41}}{2} $
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