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

The solutions to the equation (x+1)(2x-3)=5 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

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

The zeros of (x+1)(2x-3)=5 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4

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

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

Decimal Notation

x = 2.26556 \dots, x = - 1.76556 \dots

Solution steps

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

Expand

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

2 x^{2} - x - 3 = 5

Move

5

to the left side

2 x^{2} - x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 8) = 65

x_{1, 2} = \frac{- ( - 1 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 65}{2 \cdot 2}

x = \frac{- ( - 1 ) + 65}{2 \cdot 2} : \frac{1 + 65}{4}

x = \frac{- ( - 1 ) - 65}{2 \cdot 2} : \frac{1 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

x^{2} + \frac{1}{2} x - 3 = 0

x, f = x^{2} - 6 x + m - 3

x^{2} + x = \frac{19}{4}

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

y^{2} - 3 y - 10 = 0

What are the solutions to the equation (x+1)(2x-3)=5 ?
The solutions to the equation (x+1)(2x-3)=5 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4
Find the zeros of (x+1)(2x-3)=5
The zeros of (x+1)(2x-3)=5 are x=(1+sqrt(65))/4 ,x=(1-sqrt(65))/4