What are the solutions to the equation x^2+4x=285 ?

The solutions to the equation x^2+4x=285 are x=15,x=-19

Find the zeros of x^2+4x=285

The zeros of x^2+4x=285 are x=15,x=-19

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x^{2} + 4 x = 285

x = 15, x = - 19

Solution steps

x^{2} + 4 x = 285

Move

285

to the left side

x^{2} + 4 x - 285 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 285 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 285) = 34

x_{1, 2} = \frac{- 4 \pm 34}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 34}{2 \cdot 1}, x_{2} = \frac{- 4 - 34}{2 \cdot 1}

x = \frac{- 4 + 34}{2 \cdot 1} : 15

x = \frac{- 4 - 34}{2 \cdot 1} : - 19

The solutions to the quadratic equation are:

x = 15, x = - 19

(\frac{5 s}{6 t})^{- 2}

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2 x^{2} - 8 = 31 x - 2 x^{2}

3 x^{2} + 16 x - 12 < 0

What are the solutions to the equation x^2+4x=285 ?
The solutions to the equation x^2+4x=285 are x=15,x=-19
Find the zeros of x^2+4x=285
The zeros of x^2+4x=285 are x=15,x=-19