Solve Using the Quadratic Formula 16x^2-49=0

$16 x^{2} - 49 = 0$

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 $a = 16$ , $b = 0$ , and $c = - 49$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (16 \cdot - 49)}}{2 \cdot 16}$

Step 3

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 16 \cdot - 49}}{2 \cdot 16}$

Multiply $- 4 \cdot 16 \cdot - 49$ .

Multiply $- 4$ by $16$ .

$x = \frac{0 \pm \sqrt{0 - 64 \cdot - 49}}{2 \cdot 16}$

Multiply $- 64$ by $- 49$ .

$x = \frac{0 \pm \sqrt{0 + 3136}}{2 \cdot 16}$

Add $0$ and $3136$ .

$x = \frac{0 \pm \sqrt{3136}}{2 \cdot 16}$

Rewrite $3136$ as $56^{2}$ .

$x = \frac{0 \pm \sqrt{56^{2}}}{2 \cdot 16}$

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

$x = \frac{0 \pm 56}{2 \cdot 16}$

Multiply $2$ by $16$ .

$x = \frac{0 \pm 56}{32}$

Simplify $\frac{0 \pm 56}{32}$ .

$x = \frac{\pm 7}{4}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{7}{4}, - \frac{7}{4}$