What are the solutions to the equation (x+2)(x-5)=30 ?

The solutions to the equation (x+2)(x-5)=30 are x=8,x=-5

Find the zeros of (x+2)(x-5)=30

The zeros of (x+2)(x-5)=30 are x=8,x=-5

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(x + 2) (x - 5) = 30

x = 8, x = - 5

Solution steps

(x + 2) (x - 5) = 30

Expand

(x + 2) (x - 5) : x^{2} - 3 x - 10

x^{2} - 3 x - 10 = 30

Move

30

to the left side

x^{2} - 3 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 40) = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

x = \frac{- ( - 3 ) + 13}{2 \cdot 1} : 8

x = \frac{- ( - 3 ) - 13}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 8, x = - 5

(3 x - 2)^{3} < 1

5 - 3 x \leq - 1

\frac{9}{4} - \frac{9}{2}

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

6 \cdot 1 0^{- 5}

What are the solutions to the equation (x+2)(x-5)=30 ?
The solutions to the equation (x+2)(x-5)=30 are x=8,x=-5
Find the zeros of (x+2)(x-5)=30
The zeros of (x+2)(x-5)=30 are x=8,x=-5