What are the solutions to the equation 48=(2x+6)(2x+4) ?

The solutions to the equation 48=(2x+6)(2x+4) are x=1,x=-6

Find the zeros of 48=(2x+6)(2x+4)

The zeros of 48=(2x+6)(2x+4) are x=1,x=-6

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48 = (2 x + 6) (2 x + 4)

x = 1, x = - 6

Solution steps

48 = (2 x + 6) (2 x + 4)

Expand

(2 x + 6) (2 x + 4) : 4 x^{2} + 20 x + 24

48 = 4 x^{2} + 20 x + 24

Switch sides

4 x^{2} + 20 x + 24 = 48

Move

48

to the left side

4 x^{2} + 20 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 ( - 24 )}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 (- 24) = 28

x_{1, 2} = \frac{- 20 \pm 28}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 28}{2 \cdot 4}, x_{2} = \frac{- 20 - 28}{2 \cdot 4}

x = \frac{- 20 + 28}{2 \cdot 4} : 1

x = \frac{- 20 - 28}{2 \cdot 4} : - 6

The solutions to the quadratic equation are:

x = 1, x = - 6

x^{2} - 16 x = 21

x^{2} - 16 x = 20

x, x^{2} + 4 y^{2} - 4 x + 6 y + 20 = 0

X, Y = 2 X^{2}

- 16 x^{2} = - 25

What are the solutions to the equation 48=(2x+6)(2x+4) ?
The solutions to the equation 48=(2x+6)(2x+4) are x=1,x=-6
Find the zeros of 48=(2x+6)(2x+4)
The zeros of 48=(2x+6)(2x+4) are x=1,x=-6