What are the solutions to the equation (2x-7)(x+3)=4 ?

The solutions to the equation (2x-7)(x+3)=4 are x=(1+sqrt(201))/4 ,x=(1-sqrt(201))/4

Find the zeros of (2x-7)(x+3)=4

The zeros of (2x-7)(x+3)=4 are x=(1+sqrt(201))/4 ,x=(1-sqrt(201))/4

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(2 x - 7) (x + 3) = 4

x = \frac{1 + 201}{4}, x = \frac{1 - 201}{4}

Decimal Notation

x = 3.79436 \dots, x = - 3.29436 \dots

Solution steps

(2 x - 7) (x + 3) = 4

Expand

(2 x - 7) (x + 3) : 2 x^{2} - x - 21

2 x^{2} - x - 21 = 4

Move

4

to the left side

2 x^{2} - x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 25 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 25) = 201

x_{1, 2} = \frac{- ( - 1 ) \pm 201}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 201}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 201}{2 \cdot 2}

x = \frac{- ( - 1 ) + 201}{2 \cdot 2} : \frac{1 + 201}{4}

x = \frac{- ( - 1 ) - 201}{2 \cdot 2} : \frac{1 - 201}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 201}{4}, x = \frac{1 - 201}{4}

- 5 (x)^{2} - 5 x + 200 = 0

- 2 x^{2} + 8 = 6 x

3 x^{2} = 2 (3 x - 1)

n (n - 3) = 54

n (n - 3) = 70

What are the solutions to the equation (2x-7)(x+3)=4 ?
The solutions to the equation (2x-7)(x+3)=4 are x=(1+sqrt(201))/4 ,x=(1-sqrt(201))/4
Find the zeros of (2x-7)(x+3)=4
The zeros of (2x-7)(x+3)=4 are x=(1+sqrt(201))/4 ,x=(1-sqrt(201))/4