What are the solutions to the equation u^2-11u+28=0 ?

The solutions to the equation u^2-11u+28=0 are u=7,u=4

Find the zeros of u^2-11u+28=0

The zeros of u^2-11u+28=0 are u=7,u=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

u^{2} - 11 u + 28 = 0

u = 7, u = 4

Solution steps

u^{2} - 11 u + 28 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

u_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, u_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

u = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

u = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

u = 7, u = 4

q, m \cdot a = k \cdot \frac{q ^{2}}{r ^{2}}

2 x^{2} + 6 x + 4 = 24

9.08 = 0.001 (0.725 x^{2} + 15.417 x + 607.732)

z^{2} - 10 z + 9 = 0

y^{2} - 8 y - 16 = 0

What are the solutions to the equation u^2-11u+28=0 ?
The solutions to the equation u^2-11u+28=0 are u=7,u=4
Find the zeros of u^2-11u+28=0
The zeros of u^2-11u+28=0 are u=7,u=4