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2n^2+3n=9
We move all terms to the left:
2n^2+3n-(9)=0
a = 2; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·2·(-9)
Δ = 81
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{81}=9$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*2}=\frac{-12}{4} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*2}=\frac{6}{4} =1+1/2 $
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