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(2x+8)(2x+4)-32=165
We move all terms to the left:
(2x+8)(2x+4)-32-(165)=0
We add all the numbers together, and all the variables
(2x+8)(2x+4)-197=0
We multiply parentheses ..
(+4x^2+8x+16x+32)-197=0
We get rid of parentheses
4x^2+8x+16x+32-197=0
We add all the numbers together, and all the variables
4x^2+24x-165=0
a = 4; b = 24; c = -165;
Δ = b2-4ac
Δ = 242-4·4·(-165)
Δ = 3216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3216}=\sqrt{16*201}=\sqrt{16}*\sqrt{201}=4\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{201}}{2*4}=\frac{-24-4\sqrt{201}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{201}}{2*4}=\frac{-24+4\sqrt{201}}{8} $
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