x^2+3x=9=0

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Solution for x^2+3x=9=0 equation:



x^2+3x=9=0
We move all terms to the left:
x^2+3x-(9)=0
a = 1; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·1·(-9)
Δ = 45
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:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{5}}{2*1}=\frac{-3-3\sqrt{5}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{5}}{2*1}=\frac{-3+3\sqrt{5}}{2} $

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