Solve Using the Quadratic Formula t(6t+5)=0

$t (6 t + 5) = 0$

Step 1

Simplify the left side.

Simplify $t (6 t + 5)$ .

Simplify by multiplying through.

Apply the distributive property.

$t (6 t) + t \cdot 5 = 0$

Reorder.

Rewrite using the commutative property of multiplication.

$6 t \cdot t + t \cdot 5 = 0$

Move $5$ to the left of $t$ .

$6 t \cdot t + 5 \cdot t = 0$

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

Move $t$ .

$6 (t \cdot t) + 5 \cdot t = 0$

Multiply $t$ by $t$ .

$6 t^{2} + 5 \cdot t = 0$

$6 t^{2} + 5 t = 0$

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 $a = 6$ , $b = 5$ , and $c = 0$ into the quadratic formula and solve for $t$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (6 \cdot 0)}}{2 \cdot 6}$

Step 4

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$t = \frac{- 5 \pm \sqrt{25 - 4 \cdot 6 \cdot 0}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 0$ .

Multiply $- 4$ by $6$ .

$t = \frac{- 5 \pm \sqrt{25 - 24 \cdot 0}}{2 \cdot 6}$

Multiply $- 24$ by $0$ .

$t = \frac{- 5 \pm \sqrt{25 + 0}}{2 \cdot 6}$

Add $25$ and $0$ .

$t = \frac{- 5 \pm \sqrt{25}}{2 \cdot 6}$

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

$t = \frac{- 5 \pm \sqrt{5^{2}}}{2 \cdot 6}$

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

$t = \frac{- 5 \pm 5}{2 \cdot 6}$

Multiply $2$ by $6$ .

$t = \frac{- 5 \pm 5}{12}$

Step 5

The final answer is the combination of both solutions.

$t = 0, - \frac{5}{6}$