What are the solutions to the equation 15r^2-6r-3=-2r ?

The solutions to the equation 15r^2-6r-3=-2r are r= 3/5 ,r=-1/3

Find the zeros of 15r^2-6r-3=-2r

The zeros of 15r^2-6r-3=-2r are r= 3/5 ,r=-1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

15 r^{2} - 6 r - 3 = - 2 r

r = \frac{3}{5}, r = - \frac{1}{3}

Solution steps

15 r^{2} - 6 r - 3 = - 2 r

Move

2 r

to the left side

15 r^{2} - 4 r - 3 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 15 ( - 3 )}{2 \cdot 15}

(- 4)^{2} - 4 \cdot 15 (- 3) = 14

r_{1, 2} = \frac{- ( - 4 ) \pm 14}{2 \cdot 15}

Separate the solutions

r_{1} = \frac{- ( - 4 ) + 14}{2 \cdot 15}, r_{2} = \frac{- ( - 4 ) - 14}{2 \cdot 15}

r = \frac{- ( - 4 ) + 14}{2 \cdot 15} : \frac{3}{5}

r = \frac{- ( - 4 ) - 14}{2 \cdot 15} : - \frac{1}{3}

The solutions to the quadratic equation are:

r = \frac{3}{5}, r = - \frac{1}{3}

∣ 4 x - 8 ∣ > 5

7 + 2 x - x^{2} = 0

\frac{( 2 a ^{5} b ) ^{2}}{24 b ^{6}}

- 3 x^{2} - 15 + 12 x = 0

(x - 8) (x - 2) (x + 4) \leq 0

What are the solutions to the equation 15r^2-6r-3=-2r ?
The solutions to the equation 15r^2-6r-3=-2r are r= 3/5 ,r=-1/3
Find the zeros of 15r^2-6r-3=-2r
The zeros of 15r^2-6r-3=-2r are r= 3/5 ,r=-1/3