What are the solutions to the equation w^2+(w+3)^2=13^2 ?

The solutions to the equation w^2+(w+3)^2=13^2 are w=(-3+sqrt(329))/2 ,w=-(3+sqrt(329))/2

Find the zeros of w^2+(w+3)^2=13^2

The zeros of w^2+(w+3)^2=13^2 are w=(-3+sqrt(329))/2 ,w=-(3+sqrt(329))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

w^{2} + (w + 3)^{2} = 1 3^{2}

w = \frac{- 3 + 329}{2}, w = - \frac{3 + 329}{2}

Decimal Notation

w = 7.56917 \dots, w = - 10.56917 \dots

Solution steps

w^{2} + (w + 3)^{2} = 1 3^{2}

Expand

w^{2} + (w + 3)^{2} : 2 w^{2} + 6 w + 9

Expand

1 3^{2} : 169

2 w^{2} + 6 w + 9 = 169

Move

169

to the left side

2 w^{2} + 6 w - 160 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 160 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 160) = 2329

w_{1, 2} = \frac{- 6 \pm 2 329}{2 \cdot 2}

Separate the solutions

w_{1} = \frac{- 6 + 2 329}{2 \cdot 2}, w_{2} = \frac{- 6 - 2 329}{2 \cdot 2}

w = \frac{- 6 + 2 329}{2 \cdot 2} : \frac{- 3 + 329}{2}

w = \frac{- 6 - 2 329}{2 \cdot 2} : - \frac{3 + 329}{2}

The solutions to the quadratic equation are:

w = \frac{- 3 + 329}{2}, w = - \frac{3 + 329}{2}

x - 1 < 7

3 x^{2} = 4 x + 3

\frac{13}{7} \cdot 5

x + 25 = 8

30 = 19.6 + 1.74 x

What are the solutions to the equation w^2+(w+3)^2=13^2 ?
The solutions to the equation w^2+(w+3)^2=13^2 are w=(-3+sqrt(329))/2 ,w=-(3+sqrt(329))/2
Find the zeros of w^2+(w+3)^2=13^2
The zeros of w^2+(w+3)^2=13^2 are w=(-3+sqrt(329))/2 ,w=-(3+sqrt(329))/2