Solve x^2+3x-9=

$$x ^ { 2 } + 3 x - 9 =$$

Factor

$$\left(x-\frac{-3\sqrt{5}-3}{2}\right)\left(x-\frac{3\sqrt{5}-3}{2}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+3x-9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-9\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{-3±\sqrt{9+36}}{2}$$

Add 9 to 36.

$$x=\frac{-3±\sqrt{45}}{2}$$

Take the square root of 45.

$$x=\frac{-3±3\sqrt{5}}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{2} when ± is plus. Add -3 to 3\sqrt{5}.

$$x=\frac{3\sqrt{5}-3}{2}$$

Now solve the equation x=\frac{-3±3\sqrt{5}}{2} when ± is minus. Subtract 3\sqrt{5} from -3.

$$x=\frac{-3\sqrt{5}-3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+3\sqrt{5}}{2} for x_{1} and \frac{-3-3\sqrt{5}}{2} for x_{2}.

$$x^{2}+3x-9=\left(x-\frac{3\sqrt{5}-3}{2}\right)\left(x-\frac{-3\sqrt{5}-3}{2}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +3x -9 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -3 rs = -9$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{3}{2} - u s = -\frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -9

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -9$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{4} - u^2 = -9$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -9-\frac{9}{4} = -\frac{45}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{45}{4} u = \pm\sqrt{\frac{45}{4}} = \pm \frac{\sqrt{45}}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{3}{2} - \frac{\sqrt{45}}{2} = -4.854 s = -\frac{3}{2} + \frac{\sqrt{45}}{2} = 1.854$$

Evaluate

$$x^{2}+3x-9$$