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8x(3x+2)=18
We move all terms to the left:
8x(3x+2)-(18)=0
We multiply parentheses
24x^2+16x-18=0
a = 24; b = 16; c = -18;
Δ = b2-4ac
Δ = 162-4·24·(-18)
Δ = 1984
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{31}}{2*24}=\frac{-16-8\sqrt{31}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{31}}{2*24}=\frac{-16+8\sqrt{31}}{48} $
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