Solve Using the Quadratic Formula r^2+3r-4=0

$r^{2} + 3 r - 4 = 0$

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 $a = 1$ , $b = 3$ , and $c = - 4$ into the quadratic formula and solve for $r$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 4$ .

$r = \frac{- 3 \pm \sqrt{9 + 16}}{2 \cdot 1}$

Add $9$ and $16$ .

$r = \frac{- 3 \pm \sqrt{25}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$r = \frac{- 3 \pm \sqrt{5^{2}}}{2 \cdot 1}$

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

$r = \frac{- 3 \pm 5}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 3 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$r = 1, - 4$