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x2+18x-40=0
We add all the numbers together, and all the variables
x^2+18x-40=0
a = 1; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·1·(-40)
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*1}=\frac{-40}{2} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*1}=\frac{4}{2} =2 $
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