What are the solutions to the equation x^2+15x=-56 ?

The solutions to the equation x^2+15x=-56 are x=-7,x=-8

Find the zeros of x^2+15x=-56

The zeros of x^2+15x=-56 are x=-7,x=-8

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x^{2} + 15 x = - 56

x = - 7, x = - 8

Solution steps

x^{2} + 15 x = - 56

Move

56

to the left side

x^{2} + 15 x + 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 56 = 1

x_{1, 2} = \frac{- 15 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 15 + 1}{2 \cdot 1}, x_{2} = \frac{- 15 - 1}{2 \cdot 1}

x = \frac{- 15 + 1}{2 \cdot 1} : - 7

x = \frac{- 15 - 1}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 7, x = - 8

3 x \cdot 3 x

7 x - x^{2}

(x - 2) (x - 5) \leq 0

x^{3} - 3 x^{2} - 9 x + 27

∣ 2 y - 7 ∣ \geq 10

What are the solutions to the equation x^2+15x=-56 ?
The solutions to the equation x^2+15x=-56 are x=-7,x=-8
Find the zeros of x^2+15x=-56
The zeros of x^2+15x=-56 are x=-7,x=-8