What are the solutions to the equation (3x-2)(x+1)=8x-3 ?

The solutions to the equation (3x-2)(x+1)=8x-3 are x=(7+sqrt(37))/6 ,x=(7-sqrt(37))/6

Find the zeros of (3x-2)(x+1)=8x-3

The zeros of (3x-2)(x+1)=8x-3 are x=(7+sqrt(37))/6 ,x=(7-sqrt(37))/6

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(3 x - 2) (x + 1) = 8 x - 3

x = \frac{7 + 37}{6}, x = \frac{7 - 37}{6}

Decimal Notation

x = 2.18046 \dots, x = 0.15287 \dots

Solution steps

(3 x - 2) (x + 1) = 8 x - 3

Expand

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

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

Move

3

to the left side

3 x^{2} + x + 1 = 8 x

Move

8 x

to the left side

3 x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

(- 7)^{2} - 4 \cdot 3 \cdot 1 = 37

x_{1, 2} = \frac{- ( - 7 ) \pm 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 37}{2 \cdot 3}, x_{2} = \frac{- ( - 7 ) - 37}{2 \cdot 3}

x = \frac{- ( - 7 ) + 37}{2 \cdot 3} : \frac{7 + 37}{6}

x = \frac{- ( - 7 ) - 37}{2 \cdot 3} : \frac{7 - 37}{6}

The solutions to the quadratic equation are:

x = \frac{7 + 37}{6}, x = \frac{7 - 37}{6}

3 x^{2} + 5 x - 28 = 0

(1 + 5 x)^{2} = (4 x - 1) 2^{2} + 9 x (x - 2)

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

x, y = 3 x^{2} - 30 x + 72

- 2 x^{2} + 2 x + 13 = 0

What are the solutions to the equation (3x-2)(x+1)=8x-3 ?
The solutions to the equation (3x-2)(x+1)=8x-3 are x=(7+sqrt(37))/6 ,x=(7-sqrt(37))/6
Find the zeros of (3x-2)(x+1)=8x-3
The zeros of (3x-2)(x+1)=8x-3 are x=(7+sqrt(37))/6 ,x=(7-sqrt(37))/6