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X^2+2X-420=0
a = 1; b = 2; c = -420;
Δ = b2-4ac
Δ = 22-4·1·(-420)
Δ = 1684
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1684}=\sqrt{4*421}=\sqrt{4}*\sqrt{421}=2\sqrt{421}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{421}}{2*1}=\frac{-2-2\sqrt{421}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{421}}{2*1}=\frac{-2+2\sqrt{421}}{2} $
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