What are the solutions to the equation (x-2)(x+9)=-10 ?

The solutions to the equation (x-2)(x+9)=-10 are x=1,x=-8

Find the zeros of (x-2)(x+9)=-10

The zeros of (x-2)(x+9)=-10 are x=1,x=-8

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(x - 2) (x + 9) = - 10

x = 1, x = - 8

Solution steps

(x - 2) (x + 9) = - 10

Expand

(x - 2) (x + 9) : x^{2} + 7 x - 18

x^{2} + 7 x - 18 = - 10

Move

10

to the left side

x^{2} + 7 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 8) = 9

x_{1, 2} = \frac{- 7 \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 9}{2 \cdot 1}, x_{2} = \frac{- 7 - 9}{2 \cdot 1}

x = \frac{- 7 + 9}{2 \cdot 1} : 1

x = \frac{- 7 - 9}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 1, x = - 8

(x - 1) (x + 7) \leq 0

\frac{16 y x ^{4}}{9 x ^{8} y ^{2}}

(7 + 10) (7 - 10)

\frac{2}{3} x - \frac{1}{5} x = x - 1

x^{2} = x^{2}

What are the solutions to the equation (x-2)(x+9)=-10 ?
The solutions to the equation (x-2)(x+9)=-10 are x=1,x=-8
Find the zeros of (x-2)(x+9)=-10
The zeros of (x-2)(x+9)=-10 are x=1,x=-8