What are the solutions to the equation (7+1)(x^2+2x)=(7+3)(3x+5) ?

The solutions to the equation (7+1)(x^2+2x)=(7+3)(3x+5) are x=(7+sqrt(449))/8 ,x=(7-sqrt(449))/8

Find the zeros of (7+1)(x^2+2x)=(7+3)(3x+5)

The zeros of (7+1)(x^2+2x)=(7+3)(3x+5) are x=(7+sqrt(449))/8 ,x=(7-sqrt(449))/8

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(7 + 1) (x^{2} + 2 x) = (7 + 3) (3 x + 5)

x = \frac{7 + 449}{8}, x = \frac{7 - 449}{8}

Solution steps

(7 + 1) (x^{2} + 2 x) = (7 + 3) (3 x + 5)

Expand

(7 + 1) (x^{2} + 2 x) : 8 x^{2} + 16 x

Expand

(7 + 3) (3 x + 5) : 30 x + 50

8 x^{2} + 16 x = 30 x + 50

Move

50

to the left side

8 x^{2} + 16 x - 50 = 30 x

Move

30 x

to the left side

8 x^{2} - 14 x - 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 8 ( - 50 )}{2 \cdot 8}

(- 14)^{2} - 4 \cdot 8 (- 50) = 2449

x_{1, 2} = \frac{- ( - 14 ) \pm 2 449}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 449}{2 \cdot 8}, x_{2} = \frac{- ( - 14 ) - 2 449}{2 \cdot 8}

x = \frac{- ( - 14 ) + 2 449}{2 \cdot 8} : \frac{7 + 449}{8}

x = \frac{- ( - 14 ) - 2 449}{2 \cdot 8} : \frac{7 - 449}{8}

The solutions to the quadratic equation are:

x = \frac{7 + 449}{8}, x = \frac{7 - 449}{8}

y, f = 101 + ln (e^{2 x} + z^{3}) + x e^{z} - y^{2} z

x, x^{2} + 4 y^{2} + 4 x y + 12 = 0

\frac{1}{2} x^{2} + x + 1 = 0

y^{2} - 7 = 0

6 x^{2} + 10 x + 4 = 0

What are the solutions to the equation (7+1)(x^2+2x)=(7+3)(3x+5) ?
The solutions to the equation (7+1)(x^2+2x)=(7+3)(3x+5) are x=(7+sqrt(449))/8 ,x=(7-sqrt(449))/8
Find the zeros of (7+1)(x^2+2x)=(7+3)(3x+5)
The zeros of (7+1)(x^2+2x)=(7+3)(3x+5) are x=(7+sqrt(449))/8 ,x=(7-sqrt(449))/8