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(2x+4)(5x-20)=0
We multiply parentheses ..
(+10x^2-40x+20x-80)=0
We get rid of parentheses
10x^2-40x+20x-80=0
We add all the numbers together, and all the variables
10x^2-20x-80=0
a = 10; b = -20; c = -80;
Δ = b2-4ac
Δ = -202-4·10·(-80)
Δ = 3600
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{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*10}=\frac{-40}{20} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*10}=\frac{80}{20} =4 $
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