What are the solutions to the equation x^2+5x-(a^2+a-6)=0 ?

The solutions to the equation x^2+5x-(a^2+a-6)=0 are x=a-2,x=-a-3

Find the zeros of x^2+5x-(a^2+a-6)=0

The zeros of x^2+5x-(a^2+a-6)=0 are x=a-2,x=-a-3

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x^{2} + 5 x - (a^{2} + a - 6) = 0

x = a - 2, x = - a - 3

Solution steps

x^{2} + 5 x - (a^{2} + a - 6) = 0

Expand

x^{2} + 5 x - (a^{2} + a - 6) : x^{2} + 5 x - a^{2} - a + 6

x^{2} + 5 x - a^{2} - a + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot ( - a ^{2} - a + 6 )}{2 \cdot 1}

Simplify

5^{2} - 4 \cdot 1 \cdot (- a^{2} - a + 6) : 2 a + 1

x_{1, 2} = \frac{- 5 \pm ( 2 a + 1 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 2 a + 1}{2 \cdot 1}, x_{2} = \frac{- 5 - ( 2 a + 1 )}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = a - 2, x = - a - 3

y, x^{3} - x y^{2} + 3 y^{2} + 2 = 0

\frac{2 x ^{2} - ( k + 3 ) x - k + 5}{4} = 0

(x + 3 a) (x - 4 b) = 0

x, y^{11} + y^{1} + y = x^{2}

x, y^{2} (a^{2} - x^{2}) = a^{3} x

What are the solutions to the equation x^2+5x-(a^2+a-6)=0 ?
The solutions to the equation x^2+5x-(a^2+a-6)=0 are x=a-2,x=-a-3
Find the zeros of x^2+5x-(a^2+a-6)=0
The zeros of x^2+5x-(a^2+a-6)=0 are x=a-2,x=-a-3