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(4x+8)(2x+2)=0
We multiply parentheses ..
(+8x^2+8x+16x+16)=0
We get rid of parentheses
8x^2+8x+16x+16=0
We add all the numbers together, and all the variables
8x^2+24x+16=0
a = 8; b = 24; c = +16;
Δ = b2-4ac
Δ = 242-4·8·16
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8}{2*8}=\frac{-32}{16} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8}{2*8}=\frac{-16}{16} =-1 $
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