Solve Using the Quadratic Formula 49t^2-14t+1=0

$49 t^{2} - 14 t + 1 = 0$

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 $a = 49$ , $b = - 14$ , and $c = 1$ into the quadratic formula and solve for $t$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (49 \cdot 1)}}{2 \cdot 49}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$t = \frac{14 \pm \sqrt{196 - 4 \cdot 49 \cdot 1}}{2 \cdot 49}$

Multiply $- 4 \cdot 49 \cdot 1$ .

Multiply $- 4$ by $49$ .

$t = \frac{14 \pm \sqrt{196 - 196 \cdot 1}}{2 \cdot 49}$

Multiply $- 196$ by $1$ .

$t = \frac{14 \pm \sqrt{196 - 196}}{2 \cdot 49}$

Subtract $196$ from $196$ .

$t = \frac{14 \pm \sqrt{0}}{2 \cdot 49}$

Rewrite $0$ as $0^{2}$ .

$t = \frac{14 \pm \sqrt{0^{2}}}{2 \cdot 49}$

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

$t = \frac{14 \pm 0}{2 \cdot 49}$

$14$ plus or minus $0$ is $14$ .

$t = \frac{14}{2 \cdot 49}$

Multiply $2$ by $49$ .

$t = \frac{14}{98}$

Cancel the common factor of $14$ and $98$ .

Factor $14$ out of $14$ .

$t = \frac{14 (1)}{98}$

Cancel the common factors.

Factor $14$ out of $98$ .

$t = \frac{14 \cdot 1}{14 \cdot 7}$

Cancel the common factor.

$t = \frac{14 \cdot 1}{14 \cdot 7}$

Rewrite the expression.

$t = \frac{1}{7}$

Step 4

The final answer is the combination of both solutions.

$t = \frac{1}{7}$ Double roots