What are the solutions to the equation 1x^2-2x-15=0 ?

The solutions to the equation 1x^2-2x-15=0 are x=5,x=-3

Find the zeros of 1x^2-2x-15=0

The zeros of 1x^2-2x-15=0 are x=5,x=-3

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

x = 5, x = - 3

Solution steps

1 \cdot x^{2} - 2 x - 15 = 0

Refine

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} : 5

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

The solutions to the quadratic equation are:

x = 5, x = - 3

(x - 2)^{2} = (2 x + 3)^{2}

4 x^{2} + 6 x - 18 = 0

(w + 6) (2 w + 6) = 216

56 (x + 1) + 56 x = - 15 x (x + 1)

x^{2} - 2

What are the solutions to the equation 1x^2-2x-15=0 ?
The solutions to the equation 1x^2-2x-15=0 are x=5,x=-3
Find the zeros of 1x^2-2x-15=0
The zeros of 1x^2-2x-15=0 are x=5,x=-3