What are the solutions to the equation (x-3)(x+2)=6 ?

The solutions to the equation (x-3)(x+2)=6 are x=4,x=-3

Find the zeros of (x-3)(x+2)=6

The zeros of (x-3)(x+2)=6 are x=4,x=-3

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(x - 3) (x + 2) = 6

x = 4, x = - 3

Solution steps

(x - 3) (x + 2) = 6

Expand

(x - 3) (x + 2) : x^{2} - x - 6

x^{2} - x - 6 = 6

Move

6

to the left side

x^{2} - x - 12 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = - 3

x^{2} - 9 x y + 20 y^{2}

5 x^{2} - 35 x + 60

- 2 m + 5 = - 2 m - 5

∣ 5 x + 10 ∣ \leq 0

∣ x - 5 ∣ x < - 5

What are the solutions to the equation (x-3)(x+2)=6 ?
The solutions to the equation (x-3)(x+2)=6 are x=4,x=-3
Find the zeros of (x-3)(x+2)=6
The zeros of (x-3)(x+2)=6 are x=4,x=-3