What are the solutions to the equation 4x^2+4x-8=1 ?

The solutions to the equation 4x^2+4x-8=1 are x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2

Find the zeros of 4x^2+4x-8=1

The zeros of 4x^2+4x-8=1 are x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2

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4 x^{2} + 4 x - 8 = 1

x = \frac{- 1 + 10}{2}, x = - \frac{1 + 10}{2}

Decimal Notation

x = 1.08113 \dots, x = - 2.08113 \dots

Solution steps

4 x^{2} + 4 x - 8 = 1

Move

1

to the left side

4 x^{2} + 4 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 9 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 9) = 410

x_{1, 2} = \frac{- 4 \pm 4 10}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 4 + 4 10}{2 \cdot 4}, x_{2} = \frac{- 4 - 4 10}{2 \cdot 4}

x = \frac{- 4 + 4 10}{2 \cdot 4} : \frac{- 1 + 10}{2}

x = \frac{- 4 - 4 10}{2 \cdot 4} : - \frac{1 + 10}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 10}{2}, x = - \frac{1 + 10}{2}

3 a + 2 = 11

r^{2} + 2 r + 0.5 = 0

2 p^{2} + 4 p - 5 = 0

2^{15}

What are the solutions to the equation 4x^2+4x-8=1 ?
The solutions to the equation 4x^2+4x-8=1 are x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2
Find the zeros of 4x^2+4x-8=1
The zeros of 4x^2+4x-8=1 are x=(-1+sqrt(10))/2 ,x=-(1+sqrt(10))/2