What are the solutions to the equation (9x+3)^2-10(9x+3)-144=0 ?

The solutions to the equation (9x+3)^2-10(9x+3)-144=0 are x= 5/3 ,x=-11/9

Find the zeros of (9x+3)^2-10(9x+3)-144=0

The zeros of (9x+3)^2-10(9x+3)-144=0 are x= 5/3 ,x=-11/9

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(9 x + 3)^{2} - 10 (9 x + 3) - 144 = 0

x = \frac{5}{3}, x = - \frac{11}{9}

Decimal Notation

x = 1.66666 \dots, x = - 1.22222 \dots

Solution steps

(9 x + 3)^{2} - 10 (9 x + 3) - 144 = 0

Expand

(9 x + 3)^{2} - 10 (9 x + 3) - 144 : 81 x^{2} - 36 x - 165

81 x^{2} - 36 x - 165 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 81 ( - 165 )}{2 \cdot 81}

(- 36)^{2} - 4 \cdot 81 (- 165) = 234

x_{1, 2} = \frac{- ( - 36 ) \pm 234}{2 \cdot 81}

Separate the solutions

x_{1} = \frac{- ( - 36 ) + 234}{2 \cdot 81}, x_{2} = \frac{- ( - 36 ) - 234}{2 \cdot 81}

x = \frac{- ( - 36 ) + 234}{2 \cdot 81} : \frac{5}{3}

x = \frac{- ( - 36 ) - 234}{2 \cdot 81} : - \frac{11}{9}

The solutions to the quadratic equation are:

x = \frac{5}{3}, x = - \frac{11}{9}

(w + 1)^{2} = 20

u^{2} + 13 = 0

\frac{1}{x} \leq x

\frac{5 n}{30 m} + \frac{2 m + 4 n}{30 m}

\frac{11}{6} + \frac{9}{2}

What are the solutions to the equation (9x+3)^2-10(9x+3)-144=0 ?
The solutions to the equation (9x+3)^2-10(9x+3)-144=0 are x= 5/3 ,x=-11/9
Find the zeros of (9x+3)^2-10(9x+3)-144=0
The zeros of (9x+3)^2-10(9x+3)-144=0 are x= 5/3 ,x=-11/9