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9x^2+36x-11=0
a = 9; b = 36; c = -11;
Δ = b2-4ac
Δ = 362-4·9·(-11)
Δ = 1692
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1692}=\sqrt{36*47}=\sqrt{36}*\sqrt{47}=6\sqrt{47}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{47}}{2*9}=\frac{-36-6\sqrt{47}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{47}}{2*9}=\frac{-36+6\sqrt{47}}{18} $
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