What are the solutions to the equation (3x+4)^2+9(3x+4)-10=0 ?

The solutions to the equation (3x+4)^2+9(3x+4)-10=0 are x=-1,x=-14/3

Find the zeros of (3x+4)^2+9(3x+4)-10=0

The zeros of (3x+4)^2+9(3x+4)-10=0 are x=-1,x=-14/3

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(3 x + 4)^{2} + 9 (3 x + 4) - 10 = 0

x = - 1, x = - \frac{14}{3}

Decimal Notation

x = - 1, x = - 4.66666 \dots

Solution steps

(3 x + 4)^{2} + 9 (3 x + 4) - 10 = 0

Expand

(3 x + 4)^{2} + 9 (3 x + 4) - 10 : 9 x^{2} + 51 x + 42

9 x^{2} + 51 x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 51 \pm 5 1 ^{2} - 4 \cdot 9 \cdot 42}{2 \cdot 9}

5 1^{2} - 4 \cdot 9 \cdot 42 = 33

x_{1, 2} = \frac{- 51 \pm 33}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 51 + 33}{2 \cdot 9}, x_{2} = \frac{- 51 - 33}{2 \cdot 9}

x = \frac{- 51 + 33}{2 \cdot 9} : - 1

x = \frac{- 51 - 33}{2 \cdot 9} : - \frac{14}{3}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{14}{3}

∣ x + 2 ∣ \leq 5

24 x^{5} y^{2} - 16 x y

- \frac{2 2}{3} x - 3 \leq \frac{4 3}{2}

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

5 x + 8 - 7 x = - 4 x + 1

What are the solutions to the equation (3x+4)^2+9(3x+4)-10=0 ?
The solutions to the equation (3x+4)^2+9(3x+4)-10=0 are x=-1,x=-14/3
Find the zeros of (3x+4)^2+9(3x+4)-10=0
The zeros of (3x+4)^2+9(3x+4)-10=0 are x=-1,x=-14/3