Solve Using the Quadratic Formula x^2+4x-9=x+1

$x^{2} + 4 x - 9 = x + 1$

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 $x$ from both sides of the equation.

$x^{2} + 4 x - 9 - x = 1$

Subtract $1$ from both sides of the equation.

$x^{2} + 4 x - 9 - x - 1 = 0$

Simplify $x^{2} + 4 x - 9 - x - 1$ .

Subtract $x$ from $4 x$ .

$x^{2} + 3 x - 9 - 1 = 0$

Subtract $1$ from $- 9$ .

$x^{2} + 3 x - 10 = 0$

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 $a = 1$ , $b = 3$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 10)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 10}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 10$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 10}}{2 \cdot 1}$

Multiply $- 4$ by $- 10$ .

$x = \frac{- 3 \pm \sqrt{9 + 40}}{2 \cdot 1}$

Add $9$ and $40$ .

$x = \frac{- 3 \pm \sqrt{49}}{2 \cdot 1}$

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

$x = \frac{- 3 \pm \sqrt{7^{2}}}{2 \cdot 1}$

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

$x = \frac{- 3 \pm 7}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm 7}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 2, - 5$