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

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

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

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

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

x = \frac{3}{2}, x = - 4

Decimal Notation

x = 1.5, x = - 4

Solution steps

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

Expand

(x + 5) (2 x - 3) : 2 x^{2} + 7 x - 15

2 x^{2} + 7 x - 15 = 2 x - 3

Move

3

to the left side

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

Move

2 x

to the left side

2 x^{2} + 5 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 12) = 11

x_{1, 2} = \frac{- 5 \pm 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 11}{2 \cdot 2}, x_{2} = \frac{- 5 - 11}{2 \cdot 2}

x = \frac{- 5 + 11}{2 \cdot 2} : \frac{3}{2}

x = \frac{- 5 - 11}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - 4

x, 4 y^{2} + 9 x^{2} - 24 y - 72 + 144 = 0

- x^{2} + 4.6 x - 5.29 = 0

X^{2} - 6 X + 1 = 0

x, 6 x^{2} y + 5 y^{3} + 3 x^{2} = 12 - x^{2} y^{2}

x^{2} + 9 x = 43.75

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