What are the solutions to the equation x^2+5x-3=-4x-3-4x-3 ?

The solutions to the equation x^2+5x-3=-4x-3-4x-3 are x=(-13+sqrt(157))/2 ,x=(-13-sqrt(157))/2

Find the zeros of x^2+5x-3=-4x-3-4x-3

The zeros of x^2+5x-3=-4x-3-4x-3 are x=(-13+sqrt(157))/2 ,x=(-13-sqrt(157))/2

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x^{2} + 5 x - 3 = - 4 x - 3 - 4 x - 3

x = \frac{- 13 + 157}{2}, x = \frac{- 13 - 157}{2}

Decimal Notation

x = - 0.23501 \dots, x = - 12.76498 \dots

Solution steps

x^{2} + 5 x - 3 = - 4 x - 3 - 4 x - 3

Refine

x^{2} + 5 x - 3 = - 8 x - 6

Move

6

to the left side

x^{2} + 5 x + 3 = - 8 x

Move

8 x

to the left side

x^{2} + 13 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 3 = 157

x_{1, 2} = \frac{- 13 \pm 157}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 157}{2 \cdot 1}, x_{2} = \frac{- 13 - 157}{2 \cdot 1}

x = \frac{- 13 + 157}{2 \cdot 1} : \frac{- 13 + 157}{2}

x = \frac{- 13 - 157}{2 \cdot 1} : \frac{- 13 - 157}{2}

The solutions to the quadratic equation are:

x = \frac{- 13 + 157}{2}, x = \frac{- 13 - 157}{2}

- 81 v^{2} + 91 = 21

4 a^{2} - 16 a + 16 = 36 a - 144

4 x^{2} = 27 x + 7

20 = - 16 t^{2} + 100 t + 5

4 w (w + 3) + 15 = 0

What are the solutions to the equation x^2+5x-3=-4x-3-4x-3 ?
The solutions to the equation x^2+5x-3=-4x-3-4x-3 are x=(-13+sqrt(157))/2 ,x=(-13-sqrt(157))/2
Find the zeros of x^2+5x-3=-4x-3-4x-3
The zeros of x^2+5x-3=-4x-3-4x-3 are x=(-13+sqrt(157))/2 ,x=(-13-sqrt(157))/2