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9x^2+24x=16
We move all terms to the left:
9x^2+24x-(16)=0
a = 9; b = 24; c = -16;
Δ = b2-4ac
Δ = 242-4·9·(-16)
Δ = 1152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{2}}{2*9}=\frac{-24-24\sqrt{2}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{2}}{2*9}=\frac{-24+24\sqrt{2}}{18} $
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