What are the solutions to the equation (x+2)(x-7)=-3x-10 ?

The solutions to the equation (x+2)(x-7)=-3x-10 are x=1+sqrt(5),x=1-sqrt(5)

Find the zeros of (x+2)(x-7)=-3x-10

The zeros of (x+2)(x-7)=-3x-10 are x=1+sqrt(5),x=1-sqrt(5)

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(x + 2) (x - 7) = - 3 x - 10

x = 1 + 5, x = 1 - 5

Decimal Notation

x = 3.23606 \dots, x = - 1.23606 \dots

Solution steps

(x + 2) (x - 7) = - 3 x - 10

Expand

(x + 2) (x - 7) : x^{2} - 5 x - 14

x^{2} - 5 x - 14 = - 3 x - 10

Move

10

to the left side

x^{2} - 5 x - 4 = - 3 x

Move

3 x

to the left side

x^{2} - 2 x - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 1 + 5, x = 1 - 5

3 \cdot 60

- 0.25 x^{3} + 2 x^{2} - 3.5 x - 3.9 = ∣ 1.7 x ∣ + 3.1

\frac{x ^{2} + 8 x + 16}{x ^{2} + 6 x + 8}

(\frac{64}{x})^{\frac{2}{3}}

7 - ∣ 3 x + 6 ∣ = 4

What are the solutions to the equation (x+2)(x-7)=-3x-10 ?
The solutions to the equation (x+2)(x-7)=-3x-10 are x=1+sqrt(5),x=1-sqrt(5)
Find the zeros of (x+2)(x-7)=-3x-10
The zeros of (x+2)(x-7)=-3x-10 are x=1+sqrt(5),x=1-sqrt(5)