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(4n-3)(n+3)=0
We multiply parentheses ..
(+4n^2+12n-3n-9)=0
We get rid of parentheses
4n^2+12n-3n-9=0
We add all the numbers together, and all the variables
4n^2+9n-9=0
a = 4; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·4·(-9)
Δ = 225
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{225}=15$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*4}=\frac{-24}{8} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*4}=\frac{6}{8} =3/4 $
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