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(2n+3)(n-1)=0
We multiply parentheses ..
(+2n^2-2n+3n-3)=0
We get rid of parentheses
2n^2-2n+3n-3=0
We add all the numbers together, and all the variables
2n^2+n-3=0
a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
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{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4} =-1+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*2}=\frac{4}{4} =1 $
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