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

The solutions to the equation (x-5)(x+2)=-6 are x=4,x=-1

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

The zeros of (x-5)(x+2)=-6 are x=4,x=-1

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(x - 5) (x + 2) = - 6

x = 4, x = - 1

Solution steps

(x - 5) (x + 2) = - 6

Expand

(x - 5) (x + 2) : x^{2} - 3 x - 10

x^{2} - 3 x - 10 = - 6

Move

6

to the left side

x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 1

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

2 x (x - 3) = 1

5 (p^{2} - 4) - 1 = 4

- 4 x^{2} + 8 x = 7

3 - 2 x^{2} = - 5

What are the solutions to the equation (x-5)(x+2)=-6 ?
The solutions to the equation (x-5)(x+2)=-6 are x=4,x=-1
Find the zeros of (x-5)(x+2)=-6
The zeros of (x-5)(x+2)=-6 are x=4,x=-1