What are the solutions to the equation (3x+2)(x-1)-13=2x(2-x) ?

The solutions to the equation (3x+2)(x-1)-13=2x(2-x) are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

Find the zeros of (3x+2)(x-1)-13=2x(2-x)

The zeros of (3x+2)(x-1)-13=2x(2-x) are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

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(3 x + 2) (x - 1) - 13 = 2 x (2 - x)

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

Solution steps

(3 x + 2) (x - 1) - 13 = 2 x (2 - x)

Expand

(3 x + 2) (x - 1) - 13 : 3 x^{2} - x - 15

Expand

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

3 x^{2} - x - 15 = 4 x - 2 x^{2}

Move

2 x^{2}

to the left side

5 x^{2} - x - 15 = 4 x

Move

4 x

to the left side

5 x^{2} - 5 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 5 ( - 15 )}{2 \cdot 5}

(- 5)^{2} - 4 \cdot 5 (- 15) = 513

x_{1, 2} = \frac{- ( - 5 ) \pm 5 13}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 5 13}{2 \cdot 5}, x_{2} = \frac{- ( - 5 ) - 5 13}{2 \cdot 5}

x = \frac{- ( - 5 ) + 5 13}{2 \cdot 5} : \frac{1 + 13}{2}

x = \frac{- ( - 5 ) - 5 13}{2 \cdot 5} : \frac{1 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

2 x^{2} + 17 x + 30

\frac{60 m ^{- 3} m ^{2}}{- 15 n ^{- 1} n ^{5}}

3 < x + 2 < - 2

- 9 a^{2} b - 3 a^{2}

\frac{3 x - 8}{11} - 4 = - 4

What are the solutions to the equation (3x+2)(x-1)-13=2x(2-x) ?
The solutions to the equation (3x+2)(x-1)-13=2x(2-x) are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2
Find the zeros of (3x+2)(x-1)-13=2x(2-x)
The zeros of (3x+2)(x-1)-13=2x(2-x) are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2