What are the solutions to the equation x^2+3x-5=-1 ?

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

Find the zeros of x^2+3x-5=-1

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

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x^{2} + 3 x - 5 = - 1

x = 1, x = - 4

Solution steps

x^{2} + 3 x - 5 = - 1

Move

1

to the left side

x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

x = \frac{- 3 + 5}{2 \cdot 1} : 1

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

(2 x + 1) = 0

\frac{12}{x - 6} + \frac{6}{2 x - 1} \geq - 1

4 ∣ x + 4 ∣ + 6 < 30

4 u - 2 = 14

x^{2} + 2 x y - 24 y^{2}

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