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(3x-4)8x=16
We move all terms to the left:
(3x-4)8x-(16)=0
We multiply parentheses
24x^2-32x-16=0
a = 24; b = -32; c = -16;
Δ = b2-4ac
Δ = -322-4·24·(-16)
Δ = 2560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{10}}{2*24}=\frac{32-16\sqrt{10}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{10}}{2*24}=\frac{32+16\sqrt{10}}{48} $
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