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(4x+2)(5x+8)=0
We multiply parentheses ..
(+20x^2+32x+10x+16)=0
We get rid of parentheses
20x^2+32x+10x+16=0
We add all the numbers together, and all the variables
20x^2+42x+16=0
a = 20; b = 42; c = +16;
Δ = b2-4ac
Δ = 422-4·20·16
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-22}{2*20}=\frac{-64}{40} =-1+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+22}{2*20}=\frac{-20}{40} =-1/2 $
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