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5x(x-4)+2(x-4)=0
We multiply parentheses
5x^2-20x+2x-8=0
We add all the numbers together, and all the variables
5x^2-18x-8=0
a = 5; b = -18; c = -8;
Δ = b2-4ac
Δ = -182-4·5·(-8)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*5}=\frac{-4}{10} =-2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*5}=\frac{40}{10} =4 $
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