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

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $49$ .

$c = \frac{0 \pm \sqrt{0 - 196 \cdot - 16}}{2 \cdot 49}$

Multiply $- 196$ by $- 16$ .

$c = \frac{0 \pm \sqrt{0 + 3136}}{2 \cdot 49}$

Add $0$ and $3136$ .

$c = \frac{0 \pm \sqrt{3136}}{2 \cdot 49}$

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

$c = \frac{0 \pm \sqrt{56^{2}}}{2 \cdot 49}$

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

$c = \frac{0 \pm 56}{2 \cdot 49}$

Multiply $2$ by $49$ .

$c = \frac{0 \pm 56}{98}$

Simplify $\frac{0 \pm 56}{98}$ .

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

Step 4

The final answer is the combination of both solutions.

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