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2(x+8)(x+12)=0
We multiply parentheses ..
2(+x^2+12x+8x+96)=0
We multiply parentheses
2x^2+24x+16x+192=0
We add all the numbers together, and all the variables
2x^2+40x+192=0
a = 2; b = 40; c = +192;
Δ = b2-4ac
Δ = 402-4·2·192
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8}{2*2}=\frac{-48}{4} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8}{2*2}=\frac{-32}{4} =-8 $
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