Solve Using the Quadratic Formula m^2+2m-3=0

$m^{2} + 2 m - 3 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = 2$ , and $c = - 3$ into the quadratic formula and solve for $m$ .

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

Step 3

Simplify.

Simplify the numerator.

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

$m = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$m = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

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

Add $4$ and $12$ .

$m = \frac{- 2 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$m = \frac{- 2 \pm \sqrt{4^{2}}}{2 \cdot 1}$

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

$m = \frac{- 2 \pm 4}{2 \cdot 1}$

Multiply $2$ by $1$ .

$m = \frac{- 2 \pm 4}{2}$

Simplify $\frac{- 2 \pm 4}{2}$ .

$m = - 1 \pm 2$

Step 4

The final answer is the combination of both solutions.

$m = 1, - 3$