What are the solutions to the equation x^2+18x=4x-49 ?

The solutions to the equation x^2+18x=4x-49 are x=-7

Find the zeros of x^2+18x=4x-49

The zeros of x^2+18x=4x-49 are x=-7

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x^{2} + 18 x = 4 x - 49

x = - 7

Solution steps

x^{2} + 18 x = 4 x - 49

Move

49

to the left side

x^{2} + 18 x + 49 = 4 x

Move

4 x

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

A, y = A x + A^{2} - B e^{- x}

x, 5 x^{2} + 2 x^{2} y + y^{2} = 8

3 x^{2} + 6 x + 8 = 0

x^{2} + 8 x + 21 = 0

r^{2} - 2 r + 3 = 0

What are the solutions to the equation x^2+18x=4x-49 ?
The solutions to the equation x^2+18x=4x-49 are x=-7
Find the zeros of x^2+18x=4x-49
The zeros of x^2+18x=4x-49 are x=-7