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2n^2-3n-4=-2
We move all terms to the left:
2n^2-3n-4-(-2)=0
We add all the numbers together, and all the variables
2n^2-3n-2=0
a = 2; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·2·(-2)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*2}=\frac{-2}{4} =-1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*2}=\frac{8}{4} =2 $
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