What are the solutions to the equation (4x+1)^2+9(4x+1)=-18 ?

The solutions to the equation (4x+1)^2+9(4x+1)=-18 are x=-1,x=-7/4

Find the zeros of (4x+1)^2+9(4x+1)=-18

The zeros of (4x+1)^2+9(4x+1)=-18 are x=-1,x=-7/4

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(4 x + 1)^{2} + 9 (4 x + 1) = - 18

x = - 1, x = - \frac{7}{4}

Decimal Notation

x = - 1, x = - 1.75

Solution steps

(4 x + 1)^{2} + 9 (4 x + 1) = - 18

Expand

(4 x + 1)^{2} + 9 (4 x + 1) : 16 x^{2} + 44 x + 10

16 x^{2} + 44 x + 10 = - 18

Move

18

to the left side

16 x^{2} + 44 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 44 \pm 4 4 ^{2} - 4 \cdot 16 \cdot 28}{2 \cdot 16}

4 4^{2} - 4 \cdot 16 \cdot 28 = 12

x_{1, 2} = \frac{- 44 \pm 12}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 44 + 12}{2 \cdot 16}, x_{2} = \frac{- 44 - 12}{2 \cdot 16}

x = \frac{- 44 + 12}{2 \cdot 16} : - 1

x = \frac{- 44 - 12}{2 \cdot 16} : - \frac{7}{4}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{7}{4}

- 5 x^{2} - 90 x = - 555

x, 5 x^{2} - 12 y^{2} + 3 x y^{2} = 10

x^{2} = - 15

(3 p + 2)^{2} = 8

a^{2} = 8

What are the solutions to the equation (4x+1)^2+9(4x+1)=-18 ?
The solutions to the equation (4x+1)^2+9(4x+1)=-18 are x=-1,x=-7/4
Find the zeros of (4x+1)^2+9(4x+1)=-18
The zeros of (4x+1)^2+9(4x+1)=-18 are x=-1,x=-7/4