What are the solutions to the equation t^2+9t-4=-5 ?

The solutions to the equation t^2+9t-4=-5 are t=(-9+sqrt(77))/2 ,t=(-9-sqrt(77))/2

Find the zeros of t^2+9t-4=-5

The zeros of t^2+9t-4=-5 are t=(-9+sqrt(77))/2 ,t=(-9-sqrt(77))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^{2} + 9 t - 4 = - 5

t = \frac{- 9 + 77}{2}, t = \frac{- 9 - 77}{2}

Decimal Notation

t = - 0.11251 \dots, t = - 8.88748 \dots

Solution steps

t^{2} + 9 t - 4 = - 5

Move

5

to the left side

t^{2} + 9 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 1 = 77

t_{1, 2} = \frac{- 9 \pm 77}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 9 + 77}{2 \cdot 1}, t_{2} = \frac{- 9 - 77}{2 \cdot 1}

t = \frac{- 9 + 77}{2 \cdot 1} : \frac{- 9 + 77}{2}

t = \frac{- 9 - 77}{2 \cdot 1} : \frac{- 9 - 77}{2}

The solutions to the quadratic equation are:

t = \frac{- 9 + 77}{2}, t = \frac{- 9 - 77}{2}

4 x^{2} - 23 x + 15

2 (- \frac{12}{13}) (- \frac{5}{13})

36 d^{3} + 24

6 x^{2} = 0

7^{3} (\frac{1}{7})^{5}

What are the solutions to the equation t^2+9t-4=-5 ?
The solutions to the equation t^2+9t-4=-5 are t=(-9+sqrt(77))/2 ,t=(-9-sqrt(77))/2
Find the zeros of t^2+9t-4=-5
The zeros of t^2+9t-4=-5 are t=(-9+sqrt(77))/2 ,t=(-9-sqrt(77))/2