What are the solutions to the equation (7x+2)^2+6(7x+2)=27 ?

The solutions to the equation (7x+2)^2+6(7x+2)=27 are x= 1/7 ,x=-11/7

Find the zeros of (7x+2)^2+6(7x+2)=27

The zeros of (7x+2)^2+6(7x+2)=27 are x= 1/7 ,x=-11/7

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(7 x + 2)^{2} + 6 (7 x + 2) = 27

x = \frac{1}{7}, x = - \frac{11}{7}

Decimal Notation

x = 0.14285 \dots, x = - 1.57142 \dots

Solution steps

(7 x + 2)^{2} + 6 (7 x + 2) = 27

Expand

(7 x + 2)^{2} + 6 (7 x + 2) : 49 x^{2} + 70 x + 16

49 x^{2} + 70 x + 16 = 27

Move

27

to the left side

49 x^{2} + 70 x - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 70 \pm 7 0 ^{2} - 4 \cdot 49 ( - 11 )}{2 \cdot 49}

7 0^{2} - 4 \cdot 49 (- 11) = 84

x_{1, 2} = \frac{- 70 \pm 84}{2 \cdot 49}

Separate the solutions

x_{1} = \frac{- 70 + 84}{2 \cdot 49}, x_{2} = \frac{- 70 - 84}{2 \cdot 49}

x = \frac{- 70 + 84}{2 \cdot 49} : \frac{1}{7}

x = \frac{- 70 - 84}{2 \cdot 49} : - \frac{11}{7}

The solutions to the quadratic equation are:

x = \frac{1}{7}, x = - \frac{11}{7}

∣ x + 5 ∣ = ∣ 7 x + 3 ∣

q^{2} - 1 = 1

∣ x - 5 ∣ = - 2

\frac{1}{\frac{1}{100} + \frac{1}{100} + \frac{1}{200} + \frac{1}{200}}

(- 2)^{3}

What are the solutions to the equation (7x+2)^2+6(7x+2)=27 ?
The solutions to the equation (7x+2)^2+6(7x+2)=27 are x= 1/7 ,x=-11/7
Find the zeros of (7x+2)^2+6(7x+2)=27
The zeros of (7x+2)^2+6(7x+2)=27 are x= 1/7 ,x=-11/7