What are the solutions to the equation (x-3)(x+5)=-15 ?

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

Find the zeros of (x-3)(x+5)=-15

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

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(x - 3) (x + 5) = - 15

x = 0, x = - 2

Solution steps

(x - 3) (x + 5) = - 15

Expand

(x - 3) (x + 5) : x^{2} + 2 x - 15

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

Move

15

to the left side

x^{2} + 2 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

x = \frac{- 2 + 2}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

x = 0, x = - 2

x^{2} - 15 x + 75 = 0

∣ x - 4 ∣ = ∣ 2 x + 3 ∣

\frac{4}{3} + 1

- 9 a^{2} - 12 ab + 15 a^{3} b^{2} - 24 a b^{3}

5 x^{2} + 45 x - 50

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