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X2+19X=20
We move all terms to the left:
X2+19X-(20)=0
We add all the numbers together, and all the variables
X^2+19X-20=0
a = 1; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·1·(-20)
Δ = 441
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{441}=21$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*1}=\frac{-40}{2} =-20 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*1}=\frac{2}{2} =1 $
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