What are the solutions to the equation t^2+7t+12=0 ?

The solutions to the equation t^2+7t+12=0 are t=-3,t=-4

Find the zeros of t^2+7t+12=0

The zeros of t^2+7t+12=0 are t=-3,t=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^{2} + 7 t + 12 = 0

t = - 3, t = - 4

Solution steps

t^{2} + 7 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

t_{1, 2} = \frac{- 7 \pm 1}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 7 + 1}{2 \cdot 1}, t_{2} = \frac{- 7 - 1}{2 \cdot 1}

t = \frac{- 7 + 1}{2 \cdot 1} : - 3

t = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

t = - 3, t = - 4

- 1 + \frac{1}{2}

5 x^{2} - 7 x = 2

\frac{42 - 62}{62}

\frac{1}{y ^{\frac{5}{4}}}

∣ x + 3 ∣ < 6

What are the solutions to the equation t^2+7t+12=0 ?
The solutions to the equation t^2+7t+12=0 are t=-3,t=-4
Find the zeros of t^2+7t+12=0
The zeros of t^2+7t+12=0 are t=-3,t=-4