What are the solutions to the equation (3x+2)(4x-5)=-11 ?

The solutions to the equation (3x+2)(4x-5)=-11 are x= 1/3 ,x= 1/4

Find the zeros of (3x+2)(4x-5)=-11

The zeros of (3x+2)(4x-5)=-11 are x= 1/3 ,x= 1/4

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(3 x + 2) (4 x - 5) = - 11

x = \frac{1}{3}, x = \frac{1}{4}

Decimal Notation

x = 0.33333 \dots, x = 0.25

Solution steps

(3 x + 2) (4 x - 5) = - 11

Expand

(3 x + 2) (4 x - 5) : 12 x^{2} - 7 x - 10

12 x^{2} - 7 x - 10 = - 11

Move

11

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 \cdot 1 = 1

x_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 12}, x_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 12}

x = \frac{- ( - 7 ) + 1}{2 \cdot 12} : \frac{1}{3}

x = \frac{- ( - 7 ) - 1}{2 \cdot 12} : \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = \frac{1}{4}

t^{2} - 15 t + 54 = 0

4 x^{2} + 8 x + 0 = 0

x^{2} = 1 - \frac{9}{25}

6 x^{2} - 7 x - 20 = 0

(x - 18)^{2} = 17

What are the solutions to the equation (3x+2)(4x-5)=-11 ?
The solutions to the equation (3x+2)(4x-5)=-11 are x= 1/3 ,x= 1/4
Find the zeros of (3x+2)(4x-5)=-11
The zeros of (3x+2)(4x-5)=-11 are x= 1/3 ,x= 1/4