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-3n^2-16n=-48-4n^2
We move all terms to the left:
-3n^2-16n-(-48-4n^2)=0
We get rid of parentheses
-3n^2+4n^2-16n+48=0
We add all the numbers together, and all the variables
n^2-16n+48=0
a = 1; b = -16; c = +48;
Δ = b2-4ac
Δ = -162-4·1·48
Δ = 64
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{64}=8$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*1}=\frac{8}{2} =4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*1}=\frac{24}{2} =12 $
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