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2x(3x+16)=8
We move all terms to the left:
2x(3x+16)-(8)=0
We multiply parentheses
6x^2+32x-8=0
a = 6; b = 32; c = -8;
Δ = b2-4ac
Δ = 322-4·6·(-8)
Δ = 1216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{19}}{2*6}=\frac{-32-8\sqrt{19}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{19}}{2*6}=\frac{-32+8\sqrt{19}}{12} $
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