What are the solutions to the equation x^2+x+53=4-13x ?

The solutions to the equation x^2+x+53=4-13x are x=-7

Find the zeros of x^2+x+53=4-13x

The zeros of x^2+x+53=4-13x are x=-7

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x^{2} + x + 53 = 4 - 13 x

x = - 7

Solution steps

x^{2} + x + 53 = 4 - 13 x

Move

13 x

to the left side

x^{2} + 14 x + 53 = 4

Move

4

to the left side

x^{2} + 14 x + 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 49}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 49 = 0

x_{1, 2} = \frac{- 14 \pm 0}{2 \cdot 1}

x = \frac{- 14}{2 \cdot 1}

\frac{- 14}{2 \cdot 1} = - 7

x = - 7

The solution to the quadratic equation is:

x = - 7

y^{2} + y - 20

(1)^{\frac{1}{2}}

- (7 y + 0.6) = 3.6 - y

(- 2 x)^{- 1} (3 x - 1)

- x = 6

What are the solutions to the equation x^2+x+53=4-13x ?
The solutions to the equation x^2+x+53=4-13x are x=-7
Find the zeros of x^2+x+53=4-13x
The zeros of x^2+x+53=4-13x are x=-7