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b^2-9b-54=0
a = 1; b = -9; c = -54;
Δ = b2-4ac
Δ = -92-4·1·(-54)
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{33}}{2*1}=\frac{9-3\sqrt{33}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{33}}{2*1}=\frac{9+3\sqrt{33}}{2} $
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