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(2x+4)(x+2)=2x+4x+16
We move all terms to the left:
(2x+4)(x+2)-(2x+4x+16)=0
We add all the numbers together, and all the variables
(2x+4)(x+2)-(6x+16)=0
We get rid of parentheses
(2x+4)(x+2)-6x-16=0
We multiply parentheses ..
(+2x^2+4x+4x+8)-6x-16=0
We get rid of parentheses
2x^2+4x+4x-6x+8-16=0
We add all the numbers together, and all the variables
2x^2+2x-8=0
a = 2; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·2·(-8)
Δ = 68
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*2}=\frac{-2-2\sqrt{17}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*2}=\frac{-2+2\sqrt{17}}{4} $
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