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

The solutions to the equation (x+a)^2-3(x+a)-18=0 are x=-a+6,x=-a-3

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

The zeros of (x+a)^2-3(x+a)-18=0 are x=-a+6,x=-a-3

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

x = - a + 6, x = - a - 3

Solution steps

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

Expand

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

x^{2} + 2 a x + a^{2} - 3 x - 3 a - 18 = 0

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 2 a - 3 ) \pm ( 2 a - 3 ) ^{2} - 4 \cdot 1 \cdot ( a ^{2} - 3 a - 18 )}{2 \cdot 1}

(2 a - 3)^{2} - 4 \cdot 1 \cdot (a^{2} - 3 a - 18) = 9

x_{1, 2} = \frac{- ( 2 a - 3 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 2 a - 3 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( 2 a - 3 ) - 9}{2 \cdot 1}

x = \frac{- ( 2 a - 3 ) + 9}{2 \cdot 1} : - a + 6

x = \frac{- ( 2 a - 3 ) - 9}{2 \cdot 1} : - a - 3

The solutions to the quadratic equation are:

x = - a + 6, x = - a - 3

x^{2} + 12 x + 61 = 0

3 x^{2} - 10 x + 11 = 0

5 - x^{2} = 0

s^{2} + 1

x^{2} - 2 + 1 = 0

What are the solutions to the equation (x+a)^2-3(x+a)-18=0 ?
The solutions to the equation (x+a)^2-3(x+a)-18=0 are x=-a+6,x=-a-3
Find the zeros of (x+a)^2-3(x+a)-18=0
The zeros of (x+a)^2-3(x+a)-18=0 are x=-a+6,x=-a-3