Solve Using the Quadratic Formula x^2+4x-9=5x+3

$x^{2} + 4 x - 9 = 5 x + 3$

Step 1

Move all terms to the left side of the equation and simplify.

Move all the expressions to the left side of the equation.

Subtract $5 x$ from both sides of the equation.

$x^{2} + 4 x - 9 - 5 x = 3$

Subtract $3$ from both sides of the equation.

$x^{2} + 4 x - 9 - 5 x - 3 = 0$

Simplify $x^{2} + 4 x - 9 - 5 x - 3$ .

Subtract $5 x$ from $4 x$ .

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

Subtract $3$ from $- 9$ .

$x^{2} - x - 12 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 1$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot - 12)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 12$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot - 12}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$x = \frac{1 \pm \sqrt{1 + 48}}{2 \cdot 1}$

Add $1$ and $48$ .

$x = \frac{1 \pm \sqrt{49}}{2 \cdot 1}$

Rewrite $49$ as $7^{2}$ .

$x = \frac{1 \pm \sqrt{7^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{1 \pm 7}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm 7}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - 3$