(x)^2+(3x)-9=0

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

(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}
$