What are the solutions to the equation 3d^2+8d=40 ?

The solutions to the equation 3d^2+8d=40 are d=(2(sqrt(34)-2))/3 ,d=-(2(2+sqrt(34)))/3

Find the zeros of 3d^2+8d=40

The zeros of 3d^2+8d=40 are d=(2(sqrt(34)-2))/3 ,d=-(2(2+sqrt(34)))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3 d^{2} + 8 d = 40

d = \frac{2 ( 34 - 2 )}{3}, d = - \frac{2 ( 2 + 34 )}{3}

Decimal Notation

d = 2.55396 \dots, d = - 5.22063 \dots

Solution steps

3 d^{2} + 8 d = 40

Move

40

to the left side

3 d^{2} + 8 d - 40 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 ( - 40 )}{2 \cdot 3}

8^{2} - 4 \cdot 3 (- 40) = 434

d_{1, 2} = \frac{- 8 \pm 4 34}{2 \cdot 3}

Separate the solutions

d_{1} = \frac{- 8 + 4 34}{2 \cdot 3}, d_{2} = \frac{- 8 - 4 34}{2 \cdot 3}

d = \frac{- 8 + 4 34}{2 \cdot 3} : \frac{2 ( 34 - 2 )}{3}

d = \frac{- 8 - 4 34}{2 \cdot 3} : - \frac{2 ( 2 + 34 )}{3}

The solutions to the quadratic equation are:

d = \frac{2 ( 34 - 2 )}{3}, d = - \frac{2 ( 2 + 34 )}{3}

2 a - a^{2} = 3

r^{2} + 9 r + 65 = - r

b^{2} + 8 = 33

2 (2 x + 1) = x (x + 5)

x^{2} + (x - 1)^{2} = 5

What are the solutions to the equation 3d^2+8d=40 ?
The solutions to the equation 3d^2+8d=40 are d=(2(sqrt(34)-2))/3 ,d=-(2(2+sqrt(34)))/3
Find the zeros of 3d^2+8d=40
The zeros of 3d^2+8d=40 are d=(2(sqrt(34)-2))/3 ,d=-(2(2+sqrt(34)))/3