What are the solutions to the equation (2x+3)(3x-4)=-5 ?

The solutions to the equation (2x+3)(3x-4)=-5 are x=1,x=-7/6

Find the zeros of (2x+3)(3x-4)=-5

The zeros of (2x+3)(3x-4)=-5 are x=1,x=-7/6

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(2 x + 3) (3 x - 4) = - 5

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

Decimal Notation

x = 1, x = - 1.16666 \dots

Solution steps

(2 x + 3) (3 x - 4) = - 5

Expand

(2 x + 3) (3 x - 4) : 6 x^{2} + x - 12

6 x^{2} + x - 12 = - 5

Move

5

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}

6 y^{2} = 19 y - 15

x, x^{2} y - 4 x = 5

a (a + 24) = 0

4 x^{2} + 20 x + 25

- 5 x^{2} - 30 x + 95 = 0

What are the solutions to the equation (2x+3)(3x-4)=-5 ?
The solutions to the equation (2x+3)(3x-4)=-5 are x=1,x=-7/6
Find the zeros of (2x+3)(3x-4)=-5
The zeros of (2x+3)(3x-4)=-5 are x=1,x=-7/6