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(x-5)(x-4)=0
We multiply parentheses ..
(+x^2-4x-5x+20)=0
We get rid of parentheses
x^2-4x-5x+20=0
We add all the numbers together, and all the variables
x^2-9x+20=0
a = 1; b = -9; c = +20;
Δ = b2-4ac
Δ = -92-4·1·20
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-1}{2*1}=\frac{8}{2} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+1}{2*1}=\frac{10}{2} =5 $
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