What are the solutions to the equation 2r^2-12r+18=0 ?

The solutions to the equation 2r^2-12r+18=0 are r=3

Find the zeros of 2r^2-12r+18=0

The zeros of 2r^2-12r+18=0 are r=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 r^{2} - 12 r + 18 = 0

r = 3

Solution steps

2 r^{2} - 12 r + 18 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 \cdot 18}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 \cdot 18 = 0

r_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 2}

r = \frac{- ( - 12 )}{2 \cdot 2}

\frac{- ( - 12 )}{2 \cdot 2} = 3

r = 3

The solution to the quadratic equation is:

r = 3

x + 2 = 2 x^{2} - 4

n x^{2} = m

x^{2} + 7 x + 1

- 2.7 t^{2} + 40 t + 6.5 = 0

1 5^{2} - 1 3^{2} = x^{2}

What are the solutions to the equation 2r^2-12r+18=0 ?
The solutions to the equation 2r^2-12r+18=0 are r=3
Find the zeros of 2r^2-12r+18=0
The zeros of 2r^2-12r+18=0 are r=3