What are the solutions to the equation (6+2x)(10+2x)-60=36 ?

The solutions to the equation (6+2x)(10+2x)-60=36 are x=1,x=-9

Find the zeros of (6+2x)(10+2x)-60=36

The zeros of (6+2x)(10+2x)-60=36 are x=1,x=-9

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(6 + 2 x) (10 + 2 x) - 60 = 36

x = 1, x = - 9

Solution steps

(6 + 2 x) (10 + 2 x) - 60 = 36

Expand

(6 + 2 x) (10 + 2 x) - 60 : 4 x^{2} + 32 x

4 x^{2} + 32 x = 36

Move

36

to the left side

4 x^{2} + 32 x - 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 4 ( - 36 )}{2 \cdot 4}

3 2^{2} - 4 \cdot 4 (- 36) = 40

x_{1, 2} = \frac{- 32 \pm 40}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 32 + 40}{2 \cdot 4}, x_{2} = \frac{- 32 - 40}{2 \cdot 4}

x = \frac{- 32 + 40}{2 \cdot 4} : 1

x = \frac{- 32 - 40}{2 \cdot 4} : - 9

The solutions to the quadratic equation are:

x = 1, x = - 9

3 a x^{2} + 2 b x + c = 0

x^{2} + 12 x + 10 = 0

2 x^{2} - 4 x - 16 = 0

x^{2} + 18 x + 81 = 121

x^{2} - 4 x + 4 = 20

What are the solutions to the equation (6+2x)(10+2x)-60=36 ?
The solutions to the equation (6+2x)(10+2x)-60=36 are x=1,x=-9
Find the zeros of (6+2x)(10+2x)-60=36
The zeros of (6+2x)(10+2x)-60=36 are x=1,x=-9