What are the solutions to the equation v^2-13v+8=-4 ?

The solutions to the equation v^2-13v+8=-4 are v=12,v=1

Find the zeros of v^2-13v+8=-4

The zeros of v^2-13v+8=-4 are v=12,v=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

v^{2} - 13 v + 8 = - 4

v = 12, v = 1

Solution steps

v^{2} - 13 v + 8 = - 4

Move

4

to the left side

v^{2} - 13 v + 12 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 12 = 11

v_{1, 2} = \frac{- ( - 13 ) \pm 11}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 13 ) + 11}{2 \cdot 1}, v_{2} = \frac{- ( - 13 ) - 11}{2 \cdot 1}

v = \frac{- ( - 13 ) + 11}{2 \cdot 1} : 12

v = \frac{- ( - 13 ) - 11}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

v = 12, v = 1

2 x^{2} + 5 x - 7

x^{2} + 8 = 4 x

k^{2} + 14 k - 101 = - 6

x^{2} + 8 = 9 x

- m^{2} - 10 m - 16 = 0

What are the solutions to the equation v^2-13v+8=-4 ?
The solutions to the equation v^2-13v+8=-4 are v=12,v=1
Find the zeros of v^2-13v+8=-4
The zeros of v^2-13v+8=-4 are v=12,v=1