What are the solutions to the equation (x+3)(x-4)=-12 ?

The solutions to the equation (x+3)(x-4)=-12 are x=1,x=0

Find the zeros of (x+3)(x-4)=-12

The zeros of (x+3)(x-4)=-12 are x=1,x=0

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(x + 3) (x - 4) = - 12

x = 1, x = 0

Solution steps

(x + 3) (x - 4) = - 12

Expand

(x + 3) (x - 4) : x^{2} - x - 12

x^{2} - x - 12 = - 12

Move

12

to the left side

x^{2} - x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot 0 = 1

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

Separate the solutions

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

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

x = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 1, x = 0

3 z^{2} + z + 2 = 0

0.75 (8 b + 4) - 1 = 4 b + 14

3 - 2 x = 7

(x + 3)^{(2)} - 2 (x + 2)^{(2)} + (2 x + 3) (2 x + 1) = 0

- 0.8 k - 11.9 \geq - 8.3

What are the solutions to the equation (x+3)(x-4)=-12 ?
The solutions to the equation (x+3)(x-4)=-12 are x=1,x=0
Find the zeros of (x+3)(x-4)=-12
The zeros of (x+3)(x-4)=-12 are x=1,x=0