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9x^2+42x-261=0
a = 9; b = 42; c = -261;
Δ = b2-4ac
Δ = 422-4·9·(-261)
Δ = 11160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11160}=\sqrt{36*310}=\sqrt{36}*\sqrt{310}=6\sqrt{310}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{310}}{2*9}=\frac{-42-6\sqrt{310}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{310}}{2*9}=\frac{-42+6\sqrt{310}}{18} $
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