What are the solutions to the equation x^2+4x=252 ?

The solutions to the equation x^2+4x=252 are x=14,x=-18

Find the zeros of x^2+4x=252

The zeros of x^2+4x=252 are x=14,x=-18

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x^{2} + 4 x = 252

x = 14, x = - 18

Solution steps

x^{2} + 4 x = 252

Move

252

to the left side

x^{2} + 4 x - 252 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 252 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 252) = 32

x_{1, 2} = \frac{- 4 \pm 32}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 32}{2 \cdot 1}, x_{2} = \frac{- 4 - 32}{2 \cdot 1}

x = \frac{- 4 + 32}{2 \cdot 1} : 14

x = \frac{- 4 - 32}{2 \cdot 1} : - 18

The solutions to the quadratic equation are:

x = 14, x = - 18

2 x^{2} - 22 x - 24 = 0

x^{2} - 5 x - 25 = 4

4 x + 12 = x^{2}

x^{2} - 5 = x + 5

x, x^{2} + 2 y^{2} + z = 16

What are the solutions to the equation x^2+4x=252 ?
The solutions to the equation x^2+4x=252 are x=14,x=-18
Find the zeros of x^2+4x=252
The zeros of x^2+4x=252 are x=14,x=-18