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

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

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

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

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

x = 2, x = - 4

Solution steps

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

Move

5

to the left side

x^{2} + 2 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = - 4

3 x = 96

2 π + \frac{π}{6}

- 8 s + 16 = 56

(35 + 2)^{2}

14 w = - 16 - 3 w^{2}

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