What are the solutions to the equation (x+2)^2-10(x+2)=-9 ?

The solutions to the equation (x+2)^2-10(x+2)=-9 are x=7,x=-1

Find the zeros of (x+2)^2-10(x+2)=-9

The zeros of (x+2)^2-10(x+2)=-9 are x=7,x=-1

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(x + 2)^{2} - 10 (x + 2) = - 9

x = 7, x = - 1

Solution steps

(x + 2)^{2} - 10 (x + 2) = - 9

Expand

(x + 2)^{2} - 10 (x + 2) : x^{2} - 6 x - 16

x^{2} - 6 x - 16 = - 9

Move

9

to the left side

x^{2} - 6 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot (- 7) = 8

x_{1, 2} = \frac{- ( - 6 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 8}{2 \cdot 1}

x = \frac{- ( - 6 ) + 8}{2 \cdot 1} : 7

x = \frac{- ( - 6 ) - 8}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 7, x = - 1

\frac{1}{2} (2) - 1

(10)^{- 2}

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

(- \frac{1}{2} + \frac{3}{2} i)^{2}

\frac{5}{21}

What are the solutions to the equation (x+2)^2-10(x+2)=-9 ?
The solutions to the equation (x+2)^2-10(x+2)=-9 are x=7,x=-1
Find the zeros of (x+2)^2-10(x+2)=-9
The zeros of (x+2)^2-10(x+2)=-9 are x=7,x=-1