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9x^2+16x-320=0
a = 9; b = 16; c = -320;
Δ = b2-4ac
Δ = 162-4·9·(-320)
Δ = 11776
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11776}=\sqrt{256*46}=\sqrt{256}*\sqrt{46}=16\sqrt{46}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{46}}{2*9}=\frac{-16-16\sqrt{46}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{46}}{2*9}=\frac{-16+16\sqrt{46}}{18} $
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