What are the solutions to the equation (X+4)^2=9(X+3) ?

The solutions to the equation (X+4)^2=9(X+3) are X=(1+3sqrt(5))/2 ,X=(1-3sqrt(5))/2

Find the zeros of (X+4)^2=9(X+3)

The zeros of (X+4)^2=9(X+3) are X=(1+3sqrt(5))/2 ,X=(1-3sqrt(5))/2

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(X + 4)^{2} = 9 (X + 3)

X = \frac{1 + 3 5}{2}, X = \frac{1 - 3 5}{2}

Decimal Notation

X = 3.85410 \dots, X = - 2.85410 \dots

Solution steps

(X + 4)^{2} = 9 (X + 3)

Expand

(X + 4)^{2} : X^{2} + 8 X + 16

Expand

9 (X + 3) : 9 X + 27

X^{2} + 8 X + 16 = 9 X + 27

Move

27

to the left side

X^{2} + 8 X - 11 = 9 X

Move

9 X

to the left side

X^{2} - X - 11 = 0

Solve with the quadratic formula

X_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 11) = 35

X_{1, 2} = \frac{- ( - 1 ) \pm 3 5}{2 \cdot 1}

Separate the solutions

X_{1} = \frac{- ( - 1 ) + 3 5}{2 \cdot 1}, X_{2} = \frac{- ( - 1 ) - 3 5}{2 \cdot 1}

X = \frac{- ( - 1 ) + 3 5}{2 \cdot 1} : \frac{1 + 3 5}{2}

X = \frac{- ( - 1 ) - 3 5}{2 \cdot 1} : \frac{1 - 3 5}{2}

The solutions to the quadratic equation are:

X = \frac{1 + 3 5}{2}, X = \frac{1 - 3 5}{2}

x, 12 x^{2} + 76 x - 115 = 0

2 x^{2} - 8 x - 7

- 2 x^{2} + 5 x - 1

x^{2} = - x^{2} + 6

y, x y^{2} - 9 x - y - 1 = 0

What are the solutions to the equation (X+4)^2=9(X+3) ?
The solutions to the equation (X+4)^2=9(X+3) are X=(1+3sqrt(5))/2 ,X=(1-3sqrt(5))/2
Find the zeros of (X+4)^2=9(X+3)
The zeros of (X+4)^2=9(X+3) are X=(1+3sqrt(5))/2 ,X=(1-3sqrt(5))/2