What are the solutions to the equation (x-3)(x+2)=14 ?

The solutions to the equation (x-3)(x+2)=14 are x=5,x=-4

Find the zeros of (x-3)(x+2)=14

The zeros of (x-3)(x+2)=14 are x=5,x=-4

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(x - 3) (x + 2) = 14

x = 5, x = - 4

Solution steps

(x - 3) (x + 2) = 14

Expand

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

x^{2} - x - 6 = 14

Move

14

to the left side

x^{2} - x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 20) = 9

x_{1, 2} = \frac{- ( - 1 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 9}{2 \cdot 1}

x = \frac{- ( - 1 ) + 9}{2 \cdot 1} : 5

x = \frac{- ( - 1 ) - 9}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 5, x = - 4

(2 x + 5)^{2} = - 4

(x - 4) (2 x - 3) = 12

4 x^{2} = 24 x - 35

x, f = x^{2} + 2 x y + y^{2}

y, y^{2} - 2 x y^{2} = x^{3} y + 5 x^{2} y^{2} - y

What are the solutions to the equation (x-3)(x+2)=14 ?
The solutions to the equation (x-3)(x+2)=14 are x=5,x=-4
Find the zeros of (x-3)(x+2)=14
The zeros of (x-3)(x+2)=14 are x=5,x=-4