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(2x)(2x+4)=24
We move all terms to the left:
(2x)(2x+4)-(24)=0
We multiply parentheses
4x^2+8x-24=0
a = 4; b = 8; c = -24;
Δ = b2-4ac
Δ = 82-4·4·(-24)
Δ = 448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{7}}{2*4}=\frac{-8-8\sqrt{7}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{7}}{2*4}=\frac{-8+8\sqrt{7}}{8} $
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