What are the solutions to the equation (x+a)^2-(x+a)-12=0 ?

The solutions to the equation (x+a)^2-(x+a)-12=0 are x=-a+4,x=-a-3

Find the zeros of (x+a)^2-(x+a)-12=0

The zeros of (x+a)^2-(x+a)-12=0 are x=-a+4,x=-a-3

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(x + a)^{2} - (x + a) - 12 = 0

x = - a + 4, x = - a - 3

Solution steps

(x + a)^{2} - (x + a) - 12 = 0

Expand

(x + a)^{2} - (x + a) - 12 : x^{2} + 2 a x + a^{2} - x - a - 12

x^{2} + 2 a x + a^{2} - x - a - 12 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (2 a - 1) x + a^{2} - a - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 2 a - 1 ) \pm ( 2 a - 1 ) ^{2} - 4 \cdot 1 \cdot ( a ^{2} - a - 12 )}{2 \cdot 1}

(2 a - 1)^{2} - 4 \cdot 1 \cdot (a^{2} - a - 12) = 7

x_{1, 2} = \frac{- ( 2 a - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 2 a - 1 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( 2 a - 1 ) - 7}{2 \cdot 1}

x = \frac{- ( 2 a - 1 ) + 7}{2 \cdot 1} : - a + 4

x = \frac{- ( 2 a - 1 ) - 7}{2 \cdot 1} : - a - 3

The solutions to the quadratic equation are:

x = - a + 4, x = - a - 3

b^{2} = 144

g (n) = n^{2} + 3, g (8)

4 x^{2} - 8 x - 23 = 0

x^{2} + 4 x - 117 = 0

0 = 3 x^{2} - 24 x + 45

What are the solutions to the equation (x+a)^2-(x+a)-12=0 ?
The solutions to the equation (x+a)^2-(x+a)-12=0 are x=-a+4,x=-a-3
Find the zeros of (x+a)^2-(x+a)-12=0
The zeros of (x+a)^2-(x+a)-12=0 are x=-a+4,x=-a-3