What are the solutions to the equation (x+3)^2-3(x+3)-4=0 ?

The solutions to the equation (x+3)^2-3(x+3)-4=0 are x=1,x=-4

Find the zeros of (x+3)^2-3(x+3)-4=0

The zeros of (x+3)^2-3(x+3)-4=0 are x=1,x=-4

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(x + 3)^{2} - 3 (x + 3) - 4 = 0

x = 1, x = - 4

Solution steps

(x + 3)^{2} - 3 (x + 3) - 4 = 0

Expand

(x + 3)^{2} - 3 (x + 3) - 4 : x^{2} + 3 x - 4

x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

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

x = \frac{- 3 + 5}{2 \cdot 1} : 1

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

0 = - 4.9 x^{2} + 5

x^{2} - 5 x = 36

6 \geq 10

ln (1 + 2)

\frac{3 x + 4}{2} + 1 = \frac{2 x + 4}{8}

What are the solutions to the equation (x+3)^2-3(x+3)-4=0 ?
The solutions to the equation (x+3)^2-3(x+3)-4=0 are x=1,x=-4
Find the zeros of (x+3)^2-3(x+3)-4=0
The zeros of (x+3)^2-3(x+3)-4=0 are x=1,x=-4