9x^2+36x-405=0

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Solution for 9x^2+36x-405=0 equation: 



9x^2+36x-405=0

a = 9; b = 36; c = -405;
Δ = b2-4ac
Δ = 362-4·9·(-405)
Δ = 15876
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15876}=126$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-126}{2*9}=\frac{-162}{18}
=-9
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+126}{2*9}=\frac{90}{18}
=5
$

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