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n2-3n+8n=4
We move all terms to the left:
n2-3n+8n-(4)=0
We add all the numbers together, and all the variables
n^2+5n-4=0
a = 1; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·1·(-4)
Δ = 41
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{41}}{2*1}=\frac{-5-\sqrt{41}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*1}=\frac{-5+\sqrt{41}}{2} $
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