What are the solutions to the equation (x+1)^2-3(x+1)-18=0 ?

The solutions to the equation (x+1)^2-3(x+1)-18=0 are x=5,x=-4

Find the zeros of (x+1)^2-3(x+1)-18=0

The zeros of (x+1)^2-3(x+1)-18=0 are x=5,x=-4

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(x + 1)^{2} - 3 (x + 1) - 18 = 0

x = 5, x = - 4

Solution steps

(x + 1)^{2} - 3 (x + 1) - 18 = 0

Expand

(x + 1)^{2} - 3 (x + 1) - 18 : x^{2} - x - 20

x^{2} - x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 20) = 9

x_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 1}

x = \frac{- ( - 1 ) + 9}{2 \cdot 1} : 5

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

The solutions to the quadratic equation are:

x = 5, x = - 4

- 2 \cdot lo g_{2} (2) - 2 \cdot lo g_{2} (2)

(\frac{2 a}{3 b ^{2}})^{- 3}

4 (p - 6) = 36

x^{2} + 16 = 44

x^{3} + 6 x^{2}

What are the solutions to the equation (x+1)^2-3(x+1)-18=0 ?
The solutions to the equation (x+1)^2-3(x+1)-18=0 are x=5,x=-4
Find the zeros of (x+1)^2-3(x+1)-18=0
The zeros of (x+1)^2-3(x+1)-18=0 are x=5,x=-4