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2(x+4)(x+4)-8=0
We multiply parentheses ..
2(+x^2+4x+4x+16)-8=0
We multiply parentheses
2x^2+8x+8x+32-8=0
We add all the numbers together, and all the variables
2x^2+16x+24=0
a = 2; b = 16; c = +24;
Δ = b2-4ac
Δ = 162-4·2·24
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*2}=\frac{-24}{4} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*2}=\frac{-8}{4} =-2 $
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