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x^2+41x-462=0
a = 1; b = 41; c = -462;
Δ = b2-4ac
Δ = 412-4·1·(-462)
Δ = 3529
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{-(41)-\sqrt{3529}}{2*1}=\frac{-41-\sqrt{3529}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{3529}}{2*1}=\frac{-41+\sqrt{3529}}{2} $
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