What are the solutions to the equation (2x-1)(x+1)=6 ?

The solutions to the equation (2x-1)(x+1)=6 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4

Find the zeros of (2x-1)(x+1)=6

The zeros of (2x-1)(x+1)=6 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4

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(2 x - 1) (x + 1) = 6

x = \frac{- 1 + 57}{4}, x = \frac{- 1 - 57}{4}

Decimal Notation

x = 1.63745 \dots, x = - 2.13745 \dots

Solution steps

(2 x - 1) (x + 1) = 6

Expand

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

2 x^{2} + x - 1 = 6

Move

6

to the left side

2 x^{2} + x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 7) = 57

x_{1, 2} = \frac{- 1 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 57}{2 \cdot 2}, x_{2} = \frac{- 1 - 57}{2 \cdot 2}

x = \frac{- 1 + 57}{2 \cdot 2} : \frac{- 1 + 57}{4}

x = \frac{- 1 - 57}{2 \cdot 2} : \frac{- 1 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 57}{4}, x = \frac{- 1 - 57}{4}

1000 - \frac{6250}{313} + c = 50

\frac{2 Q ^{3} P}{Q P ^{- 3}}

250 x^{3} + 128 y^{3}

10 x^{2} + 5 x^{\frac{5}{2}}

8 + 3

What are the solutions to the equation (2x-1)(x+1)=6 ?
The solutions to the equation (2x-1)(x+1)=6 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4
Find the zeros of (2x-1)(x+1)=6
The zeros of (2x-1)(x+1)=6 are x=(-1+sqrt(57))/4 ,x=(-1-sqrt(57))/4