What are the solutions to the equation (11x+13)^2-4(11x+13)-21=0 ?

The solutions to the equation (11x+13)^2-4(11x+13)-21=0 are x=-6/11 ,x=-16/11

Find the zeros of (11x+13)^2-4(11x+13)-21=0

The zeros of (11x+13)^2-4(11x+13)-21=0 are x=-6/11 ,x=-16/11

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(11 x + 13)^{2} - 4 (11 x + 13) - 21 = 0

x = - \frac{6}{11}, x = - \frac{16}{11}

Solution steps

(11 x + 13)^{2} - 4 (11 x + 13) - 21 = 0

Expand

(11 x + 13)^{2} - 4 (11 x + 13) - 21 : 121 x^{2} + 242 x + 96

121 x^{2} + 242 x + 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 242 \pm 24 2 ^{2} - 4 \cdot 121 \cdot 96}{2 \cdot 121}

24 2^{2} - 4 \cdot 121 \cdot 96 = 110

x_{1, 2} = \frac{- 242 \pm 110}{2 \cdot 121}

Separate the solutions

x_{1} = \frac{- 242 + 110}{2 \cdot 121}, x_{2} = \frac{- 242 - 110}{2 \cdot 121}

x = \frac{- 242 + 110}{2 \cdot 121} : - \frac{6}{11}

x = \frac{- 242 - 110}{2 \cdot 121} : - \frac{16}{11}

The solutions to the quadratic equation are:

x = - \frac{6}{11}, x = - \frac{16}{11}

(5 - x) (2 x + 3) = 0

3 x^{2} - 132 = 0

(3 - x) (1 + 2 x) = - 9

(x + 3)^{2} + (x + 2)^{2} = 9

4 = x^{2} + 10 x

What are the solutions to the equation (11x+13)^2-4(11x+13)-21=0 ?
The solutions to the equation (11x+13)^2-4(11x+13)-21=0 are x=-6/11 ,x=-16/11
Find the zeros of (11x+13)^2-4(11x+13)-21=0
The zeros of (11x+13)^2-4(11x+13)-21=0 are x=-6/11 ,x=-16/11