What are the solutions to the equation 2(x+1)^2=8-3x ?

The solutions to the equation 2(x+1)^2=8-3x are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

Find the zeros of 2(x+1)^2=8-3x

The zeros of 2(x+1)^2=8-3x are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4

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2 (x + 1)^{2} = 8 - 3 x

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

Decimal Notation

x = 0.71221 \dots, x = - 4.21221 \dots

Solution steps

2 (x + 1)^{2} = 8 - 3 x

Expand

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

2 x^{2} + 4 x + 2 = 8 - 3 x

Move

3 x

to the left side

2 x^{2} + 7 x + 2 = 8

Move

8

to the left side

2 x^{2} + 7 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 6) = 97

x_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 97}{2 \cdot 2}, x_{2} = \frac{- 7 - 97}{2 \cdot 2}

x = \frac{- 7 + 97}{2 \cdot 2} : \frac{- 7 + 97}{4}

x = \frac{- 7 - 97}{2 \cdot 2} : \frac{- 7 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 97}{4}, x = \frac{- 7 - 97}{4}

\frac{x + 5}{x + 9} < 2

p - 19 \geq 47

∣ - 2 x + 5 ∣ - 4 > 3

6 x^{2} - 8 = 0

\frac{7 i}{5 - 8 i}

What are the solutions to the equation 2(x+1)^2=8-3x ?
The solutions to the equation 2(x+1)^2=8-3x are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4
Find the zeros of 2(x+1)^2=8-3x
The zeros of 2(x+1)^2=8-3x are x=(-7+sqrt(97))/4 ,x=(-7-sqrt(97))/4