What are the solutions to the equation (3x-4)^2+(3x-4)=0 ?

The solutions to the equation (3x-4)^2+(3x-4)=0 are x= 4/3 ,x=1

Find the zeros of (3x-4)^2+(3x-4)=0

The zeros of (3x-4)^2+(3x-4)=0 are x= 4/3 ,x=1

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(3 x - 4)^{2} + (3 x - 4) = 0

x = \frac{4}{3}, x = 1

Decimal Notation

x = 1.33333 \dots, x = 1

Solution steps

(3 x - 4)^{2} + (3 x - 4) = 0

Expand

(3 x - 4)^{2} + (3 x - 4) : 9 x^{2} - 21 x + 12

9 x^{2} - 21 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 9 \cdot 12}{2 \cdot 9}

(- 21)^{2} - 4 \cdot 9 \cdot 12 = 3

x_{1, 2} = \frac{- ( - 21 ) \pm 3}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 21 ) + 3}{2 \cdot 9}, x_{2} = \frac{- ( - 21 ) - 3}{2 \cdot 9}

x = \frac{- ( - 21 ) + 3}{2 \cdot 9} : \frac{4}{3}

x = \frac{- ( - 21 ) - 3}{2 \cdot 9} : 1

The solutions to the quadratic equation are:

x = \frac{4}{3}, x = 1

3 x^{2} + 25 x = - 50

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

y, y^{2} = e^{2 t} + k

0.4 x^{2} + 0.242 x - 0.042 = 0

x + 2 (x - 3)^{2} + 2 = 3 (x + 2)^{2}

What are the solutions to the equation (3x-4)^2+(3x-4)=0 ?
The solutions to the equation (3x-4)^2+(3x-4)=0 are x= 4/3 ,x=1
Find the zeros of (3x-4)^2+(3x-4)=0
The zeros of (3x-4)^2+(3x-4)=0 are x= 4/3 ,x=1