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x^2+40x+396=0
a = 1; b = 40; c = +396;
Δ = b2-4ac
Δ = 402-4·1·396
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4}{2*1}=\frac{-44}{2} =-22 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4}{2*1}=\frac{-36}{2} =-18 $
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