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9x^2-13x-96=0
a = 9; b = -13; c = -96;
Δ = b2-4ac
Δ = -132-4·9·(-96)
Δ = 3625
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3625}=\sqrt{25*145}=\sqrt{25}*\sqrt{145}=5\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5\sqrt{145}}{2*9}=\frac{13-5\sqrt{145}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5\sqrt{145}}{2*9}=\frac{13+5\sqrt{145}}{18} $
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