What are the solutions to the equation (3x+5)(2x-3)=-8 ?

The solutions to the equation (3x+5)(2x-3)=-8 are x=1,x=-7/6

Find the zeros of (3x+5)(2x-3)=-8

The zeros of (3x+5)(2x-3)=-8 are x=1,x=-7/6

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(3 x + 5) (2 x - 3) = - 8

x = 1, x = - \frac{7}{6}

Decimal Notation

x = 1, x = - 1.16666 \dots

Solution steps

(3 x + 5) (2 x - 3) = - 8

Expand

(3 x + 5) (2 x - 3) : 6 x^{2} + x - 15

6 x^{2} + x - 15 = - 8

Move

8

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 6 ( - 7 )}{2 \cdot 6}

1^{2} - 4 \cdot 6 (- 7) = 13

x_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 13}{2 \cdot 6}, x_{2} = \frac{- 1 - 13}{2 \cdot 6}

x = \frac{- 1 + 13}{2 \cdot 6} : 1

x = \frac{- 1 - 13}{2 \cdot 6} : - \frac{7}{6}

The solutions to the quadratic equation are:

x = 1, x = - \frac{7}{6}

50 = 2 x^{2} + 16 x + 32

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

36 x^{2} - 81 = 0

x^{2} = - 10 x - 9

2 z^{2} + 5 z = 1

What are the solutions to the equation (3x+5)(2x-3)=-8 ?
The solutions to the equation (3x+5)(2x-3)=-8 are x=1,x=-7/6
Find the zeros of (3x+5)(2x-3)=-8
The zeros of (3x+5)(2x-3)=-8 are x=1,x=-7/6