What are the solutions to the equation -4r^2+16r=0 ?

The solutions to the equation -4r^2+16r=0 are r=0,r=4

Find the zeros of -4r^2+16r=0

The zeros of -4r^2+16r=0 are r=0,r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 4 r^{2} + 16 r = 0

r = 0, r = 4

Solution steps

- 4 r^{2} + 16 r = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

1 6^{2} - 4 (- 4) \cdot 0 = 16

r_{1, 2} = \frac{- 16 \pm 16}{2 ( - 4 )}

Separate the solutions

r_{1} = \frac{- 16 + 16}{2 ( - 4 )}, r_{2} = \frac{- 16 - 16}{2 ( - 4 )}

r = \frac{- 16 + 16}{2 ( - 4 )} : 0

r = \frac{- 16 - 16}{2 ( - 4 )} : 4

The solutions to the quadratic equation are:

r = 0, r = 4

\frac{x}{13} > 1

56.25

\frac{5}{6} x = 10

e^{i \cdot π}

9 x^{2} - 16

What are the solutions to the equation -4r^2+16r=0 ?
The solutions to the equation -4r^2+16r=0 are r=0,r=4
Find the zeros of -4r^2+16r=0
The zeros of -4r^2+16r=0 are r=0,r=4