What are the solutions to the equation x^2+2x-15=0 ?

The solutions to the equation x^2+2x-15=0 are x=3,x=-5

Find the zeros of x^2+2x-15=0

The zeros of x^2+2x-15=0 are x=3,x=-5

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x^{2} + 2 x - 15 = 0

x = 3, x = - 5

Solution steps

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

200 = 1.1 x + 0.06 x^{2}

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

5 x^{2} + 6 x

2 x - 2 = 14

m^{2} + 18 m = - 9

What are the solutions to the equation x^2+2x-15=0 ?
The solutions to the equation x^2+2x-15=0 are x=3,x=-5
Find the zeros of x^2+2x-15=0
The zeros of x^2+2x-15=0 are x=3,x=-5