What are the solutions to the equation x^2+4x=192 ?

The solutions to the equation x^2+4x=192 are x=12,x=-16

Find the zeros of x^2+4x=192

The zeros of x^2+4x=192 are x=12,x=-16

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x^{2} + 4 x = 192

x = 12, x = - 16

Solution steps

x^{2} + 4 x = 192

Move

192

to the left side

x^{2} + 4 x - 192 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 192 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 192) = 28

x_{1, 2} = \frac{- 4 \pm 28}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 28}{2 \cdot 1}, x_{2} = \frac{- 4 - 28}{2 \cdot 1}

x = \frac{- 4 + 28}{2 \cdot 1} : 12

x = \frac{- 4 - 28}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

x = 12, x = - 16

125 m^{2} + 150 m + 45

5 \cdot 3

x^{2} + 11 x + 18 = 0

b^{2} + 12 ab + 36 a^{2}

\frac{3}{4} + 3

What are the solutions to the equation x^2+4x=192 ?
The solutions to the equation x^2+4x=192 are x=12,x=-16
Find the zeros of x^2+4x=192
The zeros of x^2+4x=192 are x=12,x=-16