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2n^2+3n-10=0
a = 2; b = 3; c = -10;
Δ = b2-4ac
Δ = 32-4·2·(-10)
Δ = 89
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{89}}{2*2}=\frac{-3-\sqrt{89}}{4} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*2}=\frac{-3+\sqrt{89}}{4} $
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