What are the solutions to the equation t^2+3t-19=-5 ?

The solutions to the equation t^2+3t-19=-5 are t=(-3+sqrt(65))/2 ,t=(-3-sqrt(65))/2

Find the zeros of t^2+3t-19=-5

The zeros of t^2+3t-19=-5 are t=(-3+sqrt(65))/2 ,t=(-3-sqrt(65))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^{2} + 3 t - 19 = - 5

t = \frac{- 3 + 65}{2}, t = \frac{- 3 - 65}{2}

Decimal

t = 2.53112 \dots, t = - 5.53112 \dots

Solution steps

t^{2} + 3 t - 19 = - 5

Move

5

to the left side

t^{2} + 3 t - 14 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 14 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 14) = 65

t_{1, 2} = \frac{- 3 \pm 65}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 65}{2 \cdot 1}, t_{2} = \frac{- 3 - 65}{2 \cdot 1}

t = \frac{- 3 + 65}{2 \cdot 1} : \frac{- 3 + 65}{2}

t = \frac{- 3 - 65}{2 \cdot 1} : \frac{- 3 - 65}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 65}{2}, t = \frac{- 3 - 65}{2}

3 z^{2} - 7 z + 3 = 6

(- 4 - x) (1 - x) + 6 = 0

9 v^{2} = - 30 v - 25

5 m^{2} + 7 = 52

54 = 6 + 6 a^{2}

What are the solutions to the equation t^2+3t-19=-5 ?
The solutions to the equation t^2+3t-19=-5 are t=(-3+sqrt(65))/2 ,t=(-3-sqrt(65))/2
Find the zeros of t^2+3t-19=-5
The zeros of t^2+3t-19=-5 are t=(-3+sqrt(65))/2 ,t=(-3-sqrt(65))/2