What are the solutions to the equation (x-4)(x+5)=-18 ?

The solutions to the equation (x-4)(x+5)=-18 are x=1,x=-2

Find the zeros of (x-4)(x+5)=-18

The zeros of (x-4)(x+5)=-18 are x=1,x=-2

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(x - 4) (x + 5) = - 18

x = 1, x = - 2

Solution steps

(x - 4) (x + 5) = - 18

Expand

(x - 4) (x + 5) : x^{2} + x - 20

x^{2} + x - 20 = - 18

Move

18

to the left side

x^{2} + x - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = - 2

5 n^{2} + 3 n - 36 = 0

3 x^{2} - 10 x - 13 = 0

3 x^{2} + 41 x + n = 0

x^{2} + x - 50 = 0

56 b^{2} + 112 b - 102 = 3 + 7 b^{2}

What are the solutions to the equation (x-4)(x+5)=-18 ?
The solutions to the equation (x-4)(x+5)=-18 are x=1,x=-2
Find the zeros of (x-4)(x+5)=-18
The zeros of (x-4)(x+5)=-18 are x=1,x=-2