What are the solutions to the equation (2x-3)(2x+3)-x(x+1)-5=0 ?

The solutions to the equation (2x-3)(2x+3)-x(x+1)-5=0 are x= 7/3 ,x=-2

Find the zeros of (2x-3)(2x+3)-x(x+1)-5=0

The zeros of (2x-3)(2x+3)-x(x+1)-5=0 are x= 7/3 ,x=-2

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(2 x - 3) (2 x + 3) - x (x + 1) - 5 = 0

x = \frac{7}{3}, x = - 2

Decimal Notation

x = 2.33333 \dots, x = - 2

Solution steps

(2 x - 3) (2 x + 3) - x (x + 1) - 5 = 0

Expand

(2 x - 3) (2 x + 3) - x (x + 1) - 5 : 3 x^{2} - x - 14

3 x^{2} - x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 14 )}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 (- 14) = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 3}

x = \frac{- ( - 1 ) + 13}{2 \cdot 3} : \frac{7}{3}

x = \frac{- ( - 1 ) - 13}{2 \cdot 3} : - 2

The solutions to the quadratic equation are:

x = \frac{7}{3}, x = - 2

- 7 y^{2} (- 8 x^{3} y^{3})

3 x^{2} + 19 x + 16 = 0

15

\frac{x - 6}{x ^{2} + 3} \leq 0

x^{2} + x - 42 \leq 0

What are the solutions to the equation (2x-3)(2x+3)-x(x+1)-5=0 ?
The solutions to the equation (2x-3)(2x+3)-x(x+1)-5=0 are x= 7/3 ,x=-2
Find the zeros of (2x-3)(2x+3)-x(x+1)-5=0
The zeros of (2x-3)(2x+3)-x(x+1)-5=0 are x= 7/3 ,x=-2