What are the solutions to the equation t^2+3t-2=0 ?

The solutions to the equation t^2+3t-2=0 are t=(-3+sqrt(17))/2 ,t=(-3-sqrt(17))/2

Find the zeros of t^2+3t-2=0

The zeros of t^2+3t-2=0 are t=(-3+sqrt(17))/2 ,t=(-3-sqrt(17))/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 - 2 = 0

t = \frac{- 3 + 17}{2}, t = \frac{- 3 - 17}{2}

Decimal Notation

t = 0.56155 \dots, t = - 3.56155 \dots

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 2) = 17

t_{1, 2} = \frac{- 3 \pm 17}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 3 + 17}{2 \cdot 1}, t_{2} = \frac{- 3 - 17}{2 \cdot 1}

t = \frac{- 3 + 17}{2 \cdot 1} : \frac{- 3 + 17}{2}

t = \frac{- 3 - 17}{2 \cdot 1} : \frac{- 3 - 17}{2}

The solutions to the quadratic equation are:

t = \frac{- 3 + 17}{2}, t = \frac{- 3 - 17}{2}

3 x^{2} - 24 x + 54 = 0

0 = 130 - 6 t - 16 t^{2}

n^{2} - 4 = 4

w \frac{5 w + 4}{3} = \frac{3 w}{2}

(k + 4)^{2} = 25

What are the solutions to the equation t^2+3t-2=0 ?
The solutions to the equation t^2+3t-2=0 are t=(-3+sqrt(17))/2 ,t=(-3-sqrt(17))/2
Find the zeros of t^2+3t-2=0
The zeros of t^2+3t-2=0 are t=(-3+sqrt(17))/2 ,t=(-3-sqrt(17))/2