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2x^2+x-420=0
a = 2; b = 1; c = -420;
Δ = b2-4ac
Δ = 12-4·2·(-420)
Δ = 3361
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{3361}}{2*2}=\frac{-1-\sqrt{3361}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3361}}{2*2}=\frac{-1+\sqrt{3361}}{4} $
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