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(x-5)(5x-4)=0
We multiply parentheses ..
(+5x^2-4x-25x+20)=0
We get rid of parentheses
5x^2-4x-25x+20=0
We add all the numbers together, and all the variables
5x^2-29x+20=0
a = 5; b = -29; c = +20;
Δ = b2-4ac
Δ = -292-4·5·20
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-21}{2*5}=\frac{8}{10} =4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+21}{2*5}=\frac{50}{10} =5 $
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