What are the solutions to the equation 3(x-1)^2=(2x+5) ?

The solutions to the equation 3(x-1)^2=(2x+5) are x=(4+sqrt(22))/3 ,x=(4-sqrt(22))/3

Find the zeros of 3(x-1)^2=(2x+5)

The zeros of 3(x-1)^2=(2x+5) are x=(4+sqrt(22))/3 ,x=(4-sqrt(22))/3

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3 (x - 1)^{2} = (2 x + 5)

x = \frac{4 + 22}{3}, x = \frac{4 - 22}{3}

Solution steps

3 (x - 1)^{2} = (2 x + 5)

Expand

3 (x - 1)^{2} : 3 x^{2} - 6 x + 3

3 x^{2} - 6 x + 3 = 2 x + 5

Move

5

to the left side

3 x^{2} - 6 x - 2 = 2 x

Move

2 x

to the left side

3 x^{2} - 8 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 (- 2) = 222

x_{1, 2} = \frac{- ( - 8 ) \pm 2 22}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 22}{2 \cdot 3}, x_{2} = \frac{- ( - 8 ) - 2 22}{2 \cdot 3}

x = \frac{- ( - 8 ) + 2 22}{2 \cdot 3} : \frac{4 + 22}{3}

x = \frac{- ( - 8 ) - 2 22}{2 \cdot 3} : \frac{4 - 22}{3}

The solutions to the quadratic equation are:

x = \frac{4 + 22}{3}, x = \frac{4 - 22}{3}

q + \frac{5}{9} = \frac{1}{6}

8 1^{- \frac{1}{2}}

(x - 4) (x - 2) = - 1

(4 x^{4})^{2}

What are the solutions to the equation 3(x-1)^2=(2x+5) ?
The solutions to the equation 3(x-1)^2=(2x+5) are x=(4+sqrt(22))/3 ,x=(4-sqrt(22))/3
Find the zeros of 3(x-1)^2=(2x+5)
The zeros of 3(x-1)^2=(2x+5) are x=(4+sqrt(22))/3 ,x=(4-sqrt(22))/3