What are the solutions to the equation 5d^2+9d+5=3d^2 ?

The solutions to the equation 5d^2+9d+5=3d^2 are d=(-9+sqrt(41))/4 ,d=(-9-sqrt(41))/4

Find the zeros of 5d^2+9d+5=3d^2

The zeros of 5d^2+9d+5=3d^2 are d=(-9+sqrt(41))/4 ,d=(-9-sqrt(41))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 d^{2} + 9 d + 5 = 3 d^{2}

d = \frac{- 9 + 41}{4}, d = \frac{- 9 - 41}{4}

Decimal Notation

d = - 0.64921 \dots, d = - 3.85078 \dots

Solution steps

5 d^{2} + 9 d + 5 = 3 d^{2}

Move

3 d^{2}

to the left side

2 d^{2} + 9 d + 5 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

9^{2} - 4 \cdot 2 \cdot 5 = 41

d_{1, 2} = \frac{- 9 \pm 41}{2 \cdot 2}

Separate the solutions

d_{1} = \frac{- 9 + 41}{2 \cdot 2}, d_{2} = \frac{- 9 - 41}{2 \cdot 2}

d = \frac{- 9 + 41}{2 \cdot 2} : \frac{- 9 + 41}{4}

d = \frac{- 9 - 41}{2 \cdot 2} : \frac{- 9 - 41}{4}

The solutions to the quadratic equation are:

d = \frac{- 9 + 41}{4}, d = \frac{- 9 - 41}{4}

- 4 z + 31 \geq 17 z + 23

y^{2} = 25 + 4 + R

\frac{1}{2} y - 7 = \frac{1}{4} y

3 x^{2} - 12 x = 69

26 = 13 x - 7 x^{2}

What are the solutions to the equation 5d^2+9d+5=3d^2 ?
The solutions to the equation 5d^2+9d+5=3d^2 are d=(-9+sqrt(41))/4 ,d=(-9-sqrt(41))/4
Find the zeros of 5d^2+9d+5=3d^2
The zeros of 5d^2+9d+5=3d^2 are d=(-9+sqrt(41))/4 ,d=(-9-sqrt(41))/4