What are the solutions to the equation 2t^2+18t=-4 ?

The solutions to the equation 2t^2+18t=-4 are t=(-9+sqrt(73))/2 ,t=-(9+sqrt(73))/2

Find the zeros of 2t^2+18t=-4

The zeros of 2t^2+18t=-4 are t=(-9+sqrt(73))/2 ,t=-(9+sqrt(73))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 t^{2} + 18 t = - 4

t = \frac{- 9 + 73}{2}, t = - \frac{9 + 73}{2}

Decimal Notation

t = - 0.22799 \dots, t = - 8.77200 \dots

Solution steps

2 t^{2} + 18 t = - 4

Move

4

to the left side

2 t^{2} + 18 t + 4 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

1 8^{2} - 4 \cdot 2 \cdot 4 = 273

t_{1, 2} = \frac{- 18 \pm 2 73}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 18 + 2 73}{2 \cdot 2}, t_{2} = \frac{- 18 - 2 73}{2 \cdot 2}

t = \frac{- 18 + 2 73}{2 \cdot 2} : \frac{- 9 + 73}{2}

t = \frac{- 18 - 2 73}{2 \cdot 2} : - \frac{9 + 73}{2}

The solutions to the quadratic equation are:

t = \frac{- 9 + 73}{2}, t = - \frac{9 + 73}{2}

16 x^{2} - 16 x + 4 = 0

(2 (- \frac{3}{2}) - 1) x^{2} + 4 x + 2 = 0

8 x^{2} + 14 x + 9 = 0

x^{2} + 24 + 144 = 0

x^{2} + x = 3 + 2 x

What are the solutions to the equation 2t^2+18t=-4 ?
The solutions to the equation 2t^2+18t=-4 are t=(-9+sqrt(73))/2 ,t=-(9+sqrt(73))/2
Find the zeros of 2t^2+18t=-4
The zeros of 2t^2+18t=-4 are t=(-9+sqrt(73))/2 ,t=-(9+sqrt(73))/2