What are the solutions to the equation (5+2x)(9+2x)-45=32 ?

The solutions to the equation (5+2x)(9+2x)-45=32 are x=1,x=-8

Find the zeros of (5+2x)(9+2x)-45=32

The zeros of (5+2x)(9+2x)-45=32 are x=1,x=-8

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(5 + 2 x) (9 + 2 x) - 45 = 32

x = 1, x = - 8

Solution steps

(5 + 2 x) (9 + 2 x) - 45 = 32

Expand

(5 + 2 x) (9 + 2 x) - 45 : 4 x^{2} + 28 x

4 x^{2} + 28 x = 32

Move

32

to the left side

4 x^{2} + 28 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 ( - 32 )}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 (- 32) = 36

x_{1, 2} = \frac{- 28 \pm 36}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 36}{2 \cdot 4}, x_{2} = \frac{- 28 - 36}{2 \cdot 4}

x = \frac{- 28 + 36}{2 \cdot 4} : 1

x = \frac{- 28 - 36}{2 \cdot 4} : - 8

The solutions to the quadratic equation are:

x = 1, x = - 8

\frac{y ^{5}}{y ^{7}}

e^{- 8}

y + 6 = 17

2 x^{3} - 2 x

(2 x + 4) (x - 3) - 5 (x - 3)

What are the solutions to the equation (5+2x)(9+2x)-45=32 ?
The solutions to the equation (5+2x)(9+2x)-45=32 are x=1,x=-8
Find the zeros of (5+2x)(9+2x)-45=32
The zeros of (5+2x)(9+2x)-45=32 are x=1,x=-8