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

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

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

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

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x^{2} = - 56 - 15 x

x = - 7, x = - 8

Solution steps

x^{2} = - 56 - 15 x

Move

15 x

to the left side

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

6 x + 3 = 2 x - 9

(0.02)^{4}

t e^{- t}

7 - 3 x \leq 4 - 9 x

c^{2} - 8 c + 15 = 0

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