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x^2+x-1350=0
a = 1; b = 1; c = -1350;
Δ = b2-4ac
Δ = 12-4·1·(-1350)
Δ = 5401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{5401}}{2*1}=\frac{-1-\sqrt{5401}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5401}}{2*1}=\frac{-1+\sqrt{5401}}{2} $
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