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9x^2+24x=41
We move all terms to the left:
9x^2+24x-(41)=0
a = 9; b = 24; c = -41;
Δ = b2-4ac
Δ = 242-4·9·(-41)
Δ = 2052
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2052}=\sqrt{36*57}=\sqrt{36}*\sqrt{57}=6\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{57}}{2*9}=\frac{-24-6\sqrt{57}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{57}}{2*9}=\frac{-24+6\sqrt{57}}{18} $
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