What are the solutions to the equation x^2+18x+54=4x+9 ?

The solutions to the equation x^2+18x+54=4x+9 are x=-5,x=-9

Find the zeros of x^2+18x+54=4x+9

The zeros of x^2+18x+54=4x+9 are x=-5,x=-9

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x^{2} + 18 x + 54 = 4 x + 9

x = - 5, x = - 9

Solution steps

x^{2} + 18 x + 54 = 4 x + 9

Move

9

to the left side

x^{2} + 18 x + 45 = 4 x

Move

4 x

to the left side

x^{2} + 14 x + 45 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 45}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 45 = 4

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

Separate the solutions

x_{1} = \frac{- 14 + 4}{2 \cdot 1}, x_{2} = \frac{- 14 - 4}{2 \cdot 1}

x = \frac{- 14 + 4}{2 \cdot 1} : - 5

x = \frac{- 14 - 4}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = - 5, x = - 9

- 57 = - (- p + 1) + 2 (6 + 8 p)

\frac{5 x ^{3} y ^{3}}{125 x ^{5} y ^{6}}

\frac{10 !}{7 !}

(x + 3) (x - 2) > 0

n + 5 \geq 6

What are the solutions to the equation x^2+18x+54=4x+9 ?
The solutions to the equation x^2+18x+54=4x+9 are x=-5,x=-9
Find the zeros of x^2+18x+54=4x+9
The zeros of x^2+18x+54=4x+9 are x=-5,x=-9