What are the solutions to the equation (2x+3)(x-4)+(2x+3)(x-6)=0 ?

The solutions to the equation (2x+3)(x-4)+(2x+3)(x-6)=0 are x=5,x=-3/2

Find the zeros of (2x+3)(x-4)+(2x+3)(x-6)=0

The zeros of (2x+3)(x-4)+(2x+3)(x-6)=0 are x=5,x=-3/2

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(2 x + 3) (x - 4) + (2 x + 3) (x - 6) = 0

x = 5, x = - \frac{3}{2}

Decimal Notation

x = 5, x = - 1.5

Solution steps

(2 x + 3) (x - 4) + (2 x + 3) (x - 6) = 0

Expand

(2 x + 3) (x - 4) + (2 x + 3) (x - 6) : 4 x^{2} - 14 x - 30

4 x^{2} - 14 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 4 ( - 30 )}{2 \cdot 4}

(- 14)^{2} - 4 \cdot 4 (- 30) = 26

x_{1, 2} = \frac{- ( - 14 ) \pm 26}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 26}{2 \cdot 4}, x_{2} = \frac{- ( - 14 ) - 26}{2 \cdot 4}

x = \frac{- ( - 14 ) + 26}{2 \cdot 4} : 5

x = \frac{- ( - 14 ) - 26}{2 \cdot 4} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 5, x = - \frac{3}{2}

6 x^{2} - 29 x = 0

(x - 1)^{2} + (7 - 2 x)^{2} = 10

(x + 5) (2 x - 3) = 2 (x + 1)

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

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

What are the solutions to the equation (2x+3)(x-4)+(2x+3)(x-6)=0 ?
The solutions to the equation (2x+3)(x-4)+(2x+3)(x-6)=0 are x=5,x=-3/2
Find the zeros of (2x+3)(x-4)+(2x+3)(x-6)=0
The zeros of (2x+3)(x-4)+(2x+3)(x-6)=0 are x=5,x=-3/2