What are the solutions to the equation (x-1)(x+2)=18 ?

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

Find the zeros of (x-1)(x+2)=18

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

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(x - 1) (x + 2) = 18

x = 4, x = - 5

Solution steps

(x - 1) (x + 2) = 18

Expand

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

x^{2} + x - 2 = 18

Move

18

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} : 4

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

The solutions to the quadratic equation are:

x = 4, x = - 5

x, (x - t)^{2} + (y + 2 t)^{2} = 169

3 (n + 1)^{2} - 7 = 20

16 z^{2} = 25

x, 5 x^{2} + 2 x y + 10 y^{2} - 12 x - 22 y + 17 = 0

- \frac{x ^{2}}{4} - \frac{4 x}{3} + 7 = 0

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