What are the solutions to the equation 9(5-4x)^2-43=-39 ?

The solutions to the equation 9(5-4x)^2-43=-39 are x= 13/12 ,x= 17/12

Find the zeros of 9(5-4x)^2-43=-39

The zeros of 9(5-4x)^2-43=-39 are x= 13/12 ,x= 17/12

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9 (5 - 4 x)^{2} - 43 = - 39

x = \frac{13}{12}, x = \frac{17}{12}

Decimal Notation

x = 1.08333 \dots, x = 1.41666 \dots

Solution steps

9 (5 - 4 x)^{2} - 43 = - 39

Move

43

to the right side

9 (5 - 4 x)^{2} = 4

Divide both sides by

9

(5 - 4 x)^{2} = \frac{4}{9}

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

5 - 4 x = \frac{4}{9} : x = \frac{13}{12}

Solve

5 - 4 x = - \frac{4}{9} : x = \frac{17}{12}

The solutions to the quadratic equation are:

x = \frac{13}{12}, x = \frac{17}{12}

5 t (6 t + 4)^{\frac{- 1}{2}} + 6 (6 t + 4)^{\frac{1}{2}}

\frac{- 6}{- 3}

0.22 m = 49.95 + 0.08 m

6 x^{2} - 3 x + 4 = 8 x + 1

4 x^{2} - 13 x - 12

What are the solutions to the equation 9(5-4x)^2-43=-39 ?
The solutions to the equation 9(5-4x)^2-43=-39 are x= 13/12 ,x= 17/12
Find the zeros of 9(5-4x)^2-43=-39
The zeros of 9(5-4x)^2-43=-39 are x= 13/12 ,x= 17/12