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2n^2-3n+2=60
We move all terms to the left:
2n^2-3n+2-(60)=0
We add all the numbers together, and all the variables
2n^2-3n-58=0
a = 2; b = -3; c = -58;
Δ = b2-4ac
Δ = -32-4·2·(-58)
Δ = 473
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{473}}{2*2}=\frac{3-\sqrt{473}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{473}}{2*2}=\frac{3+\sqrt{473}}{4} $
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