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n^2+3n-88=0
a = 1; b = 3; c = -88;
Δ = b2-4ac
Δ = 32-4·1·(-88)
Δ = 361
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{361}=19$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-19}{2*1}=\frac{-22}{2} =-11 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+19}{2*1}=\frac{16}{2} =8 $
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