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8x(3x-2)=32
We move all terms to the left:
8x(3x-2)-(32)=0
We multiply parentheses
24x^2-16x-32=0
a = 24; b = -16; c = -32;
Δ = b2-4ac
Δ = -162-4·24·(-32)
Δ = 3328
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{13}}{2*24}=\frac{16-16\sqrt{13}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{13}}{2*24}=\frac{16+16\sqrt{13}}{48} $
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