What are the solutions to the equation 4r(r-3)-3(r-3)=0 ?

The solutions to the equation 4r(r-3)-3(r-3)=0 are r=3,r= 3/4

Find the zeros of 4r(r-3)-3(r-3)=0

The zeros of 4r(r-3)-3(r-3)=0 are r=3,r= 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 r (r - 3) - 3 (r - 3) = 0

r = 3, r = \frac{3}{4}

Decimal Notation

r = 3, r = 0.75

Solution steps

4 r (r - 3) - 3 (r - 3) = 0

Expand

4 r (r - 3) - 3 (r - 3) : 4 r^{2} - 15 r + 9

4 r^{2} - 15 r + 9 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

(- 15)^{2} - 4 \cdot 4 \cdot 9 = 9

r_{1, 2} = \frac{- ( - 15 ) \pm 9}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- ( - 15 ) + 9}{2 \cdot 4}, r_{2} = \frac{- ( - 15 ) - 9}{2 \cdot 4}

r = \frac{- ( - 15 ) + 9}{2 \cdot 4} : 3

r = \frac{- ( - 15 ) - 9}{2 \cdot 4} : \frac{3}{4}

The solutions to the quadratic equation are:

r = 3, r = \frac{3}{4}

x^{2} - 18 x - 32 = 0

k^{2} + 8 k + 12 = 0

- 6 x + 5 x^{2} = - 8

x, 5 x^{2} + 5 y^{2} = 0

n, x = \frac{n ( n + 1 )}{2}

What are the solutions to the equation 4r(r-3)-3(r-3)=0 ?
The solutions to the equation 4r(r-3)-3(r-3)=0 are r=3,r= 3/4
Find the zeros of 4r(r-3)-3(r-3)=0
The zeros of 4r(r-3)-3(r-3)=0 are r=3,r= 3/4