What are the solutions to the equation x^2+3x-1=-x+4 ?

The solutions to the equation x^2+3x-1=-x+4 are x=1,x=-5

Find the zeros of x^2+3x-1=-x+4

The zeros of x^2+3x-1=-x+4 are x=1,x=-5

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x^{2} + 3 x - 1 = - x + 4

x = 1, x = - 5

Solution steps

x^{2} + 3 x - 1 = - x + 4

Move

4

to the left side

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

Move

x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

x_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 6}{2 \cdot 1}, x_{2} = \frac{- 4 - 6}{2 \cdot 1}

x = \frac{- 4 + 6}{2 \cdot 1} : 1

x = \frac{- 4 - 6}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 1, x = - 5

0 = - 16 t^{2} + 128 t + 144

x^{2} + 6 x + 73

- 150 = 90 t - 4.9 t^{2}

2 x^{2} + x = 15

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

What are the solutions to the equation x^2+3x-1=-x+4 ?
The solutions to the equation x^2+3x-1=-x+4 are x=1,x=-5
Find the zeros of x^2+3x-1=-x+4
The zeros of x^2+3x-1=-x+4 are x=1,x=-5