What are the solutions to the equation y^2+12y=17 ?

The solutions to the equation y^2+12y=17 are y=-6+sqrt(53),y=-6-sqrt(53)

Find the zeros of y^2+12y=17

The zeros of y^2+12y=17 are y=-6+sqrt(53),y=-6-sqrt(53)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{2} + 12 y = 17

y = - 6 + 53, y = - 6 - 53

Decimal Notation

y = 1.28010 \dots, y = - 13.28010 \dots

Solution steps

y^{2} + 12 y = 17

Move

17

to the left side

y^{2} + 12 y - 17 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 17 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 17) = 253

y_{1, 2} = \frac{- 12 \pm 2 53}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 12 + 2 53}{2 \cdot 1}, y_{2} = \frac{- 12 - 2 53}{2 \cdot 1}

y = \frac{- 12 + 2 53}{2 \cdot 1} : - 6 + 53

y = \frac{- 12 - 2 53}{2 \cdot 1} : - 6 - 53

The solutions to the quadratic equation are:

y = - 6 + 53, y = - 6 - 53

r^{2} - 14 r + 49

\frac{49}{24} (\frac{6}{35})

- x^{2} - 8 x - 15 = 0

4 k - 14 + 3 k = 21

x^{2} - 10 x + 43 = 0

What are the solutions to the equation y^2+12y=17 ?
The solutions to the equation y^2+12y=17 are y=-6+sqrt(53),y=-6-sqrt(53)
Find the zeros of y^2+12y=17
The zeros of y^2+12y=17 are y=-6+sqrt(53),y=-6-sqrt(53)