What are the solutions to the equation (x+3)(x-1)=12 ?

The solutions to the equation (x+3)(x-1)=12 are x=3,x=-5

Find the zeros of (x+3)(x-1)=12

The zeros of (x+3)(x-1)=12 are x=3,x=-5

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(x + 3) (x - 1) = 12

x = 3, x = - 5

Solution steps

(x + 3) (x - 1) = 12

Expand

(x + 3) (x - 1) : x^{2} + 2 x - 3

x^{2} + 2 x - 3 = 12

Move

12

to the left side

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

y, \frac{3 y ^{2}}{2} = x^{2} + x + c

13 t^{2} - 8 t + 1 = - 3 t

3 x^{2} + 7 x - 49 = 0

x^{2} + x = 342

x, (x - y)^{2} + x y = 1

What are the solutions to the equation (x+3)(x-1)=12 ?
The solutions to the equation (x+3)(x-1)=12 are x=3,x=-5
Find the zeros of (x+3)(x-1)=12
The zeros of (x+3)(x-1)=12 are x=3,x=-5