Solve Using the Quadratic Formula t(4t+1)=0

$t (4 t + 1) = 0$

Step 1

Simplify the left side.

Simplify $t (4 t + 1)$ .

Apply the distributive property.

$t (4 t) + t \cdot 1 = 0$

Simplify the expression.

Rewrite using the commutative property of multiplication.

$4 t \cdot t + t \cdot 1 = 0$

Multiply $t$ by $1$ .

$4 t \cdot t + t = 0$

Multiply $t$ by $t$ by adding the exponents.

Move $t$ .

$4 (t \cdot t) + t = 0$

Multiply $t$ by $t$ .

$4 t^{2} + t = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 4$ , $b = 1$ , and $c = 0$ into the quadratic formula and solve for $t$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (4 \cdot 0)}}{2 \cdot 4}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$t = \frac{- 1 \pm \sqrt{1 - 4 \cdot 4 \cdot 0}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 0$ .

Multiply $- 4$ by $4$ .

$t = \frac{- 1 \pm \sqrt{1 - 16 \cdot 0}}{2 \cdot 4}$

Multiply $- 16$ by $0$ .

$t = \frac{- 1 \pm \sqrt{1 + 0}}{2 \cdot 4}$

Add $1$ and $0$ .

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

Any root of $1$ is $1$ .

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

Multiply $2$ by $4$ .

$t = \frac{- 1 \pm 1}{8}$

Step 5

The final answer is the combination of both solutions.

$t = 0, - \frac{1}{4}$