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9x^2-48x-7=0
a = 9; b = -48; c = -7;
Δ = b2-4ac
Δ = -482-4·9·(-7)
Δ = 2556
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2556}=\sqrt{36*71}=\sqrt{36}*\sqrt{71}=6\sqrt{71}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-6\sqrt{71}}{2*9}=\frac{48-6\sqrt{71}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+6\sqrt{71}}{2*9}=\frac{48+6\sqrt{71}}{18} $
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