What are the solutions to the equation (2x+3)(x-1)=42 ?

The solutions to the equation (2x+3)(x-1)=42 are x= 9/2 ,x=-5

Find the zeros of (2x+3)(x-1)=42

The zeros of (2x+3)(x-1)=42 are x= 9/2 ,x=-5

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(2 x + 3) (x - 1) = 42

x = \frac{9}{2}, x = - 5

Decimal Notation

x = 4.5, x = - 5

Solution steps

(2 x + 3) (x - 1) = 42

Expand

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

2 x^{2} + x - 3 = 42

Move

42

to the left side

2 x^{2} + x - 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 45 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 45) = 19

x_{1, 2} = \frac{- 1 \pm 19}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 19}{2 \cdot 2}, x_{2} = \frac{- 1 - 19}{2 \cdot 2}

x = \frac{- 1 + 19}{2 \cdot 2} : \frac{9}{2}

x = \frac{- 1 - 19}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = \frac{9}{2}, x = - 5

(2 x + 1)^{2} = 4 x + 1

6 x^{2} + 17 x + 9 = 0

t^{2} - 82 t + 160 = 0

9 w^{2} = 18 w + 7

z, z^{2} - 6 z + 25 = 0

What are the solutions to the equation (2x+3)(x-1)=42 ?
The solutions to the equation (2x+3)(x-1)=42 are x= 9/2 ,x=-5
Find the zeros of (2x+3)(x-1)=42
The zeros of (2x+3)(x-1)=42 are x= 9/2 ,x=-5