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9x^2-68x+96=0
a = 9; b = -68; c = +96;
Δ = b2-4ac
Δ = -682-4·9·96
Δ = 1168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-4\sqrt{73}}{2*9}=\frac{68-4\sqrt{73}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{73}}{2*9}=\frac{68+4\sqrt{73}}{18} $
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