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2x(4x+8)=2
We move all terms to the left:
2x(4x+8)-(2)=0
We multiply parentheses
8x^2+16x-2=0
a = 8; b = 16; c = -2;
Δ = b2-4ac
Δ = 162-4·8·(-2)
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{5}}{2*8}=\frac{-16-8\sqrt{5}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{5}}{2*8}=\frac{-16+8\sqrt{5}}{16} $
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