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9x^2+39x-160=0
a = 9; b = 39; c = -160;
Δ = b2-4ac
Δ = 392-4·9·(-160)
Δ = 7281
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7281}=\sqrt{9*809}=\sqrt{9}*\sqrt{809}=3\sqrt{809}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3\sqrt{809}}{2*9}=\frac{-39-3\sqrt{809}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3\sqrt{809}}{2*9}=\frac{-39+3\sqrt{809}}{18} $
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