What are the solutions to the equation 2(12+2x)-((3x-2)^2)/5 =20 ?

The solutions to the equation 2(12+2x)-((3x-2)^2)/5 =20 are x=-4/9 ,x=4

Find the zeros of 2(12+2x)-((3x-2)^2)/5 =20

The zeros of 2(12+2x)-((3x-2)^2)/5 =20 are x=-4/9 ,x=4

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2 (12 + 2 x) - \frac{( 3 x - 2 ) ^{2}}{5} = 20

x = - \frac{4}{9}, x = 4

Decimal Notation

x = - 0.44444 \dots, x = 4

Solution steps

2 (12 + 2 x) - \frac{( 3 x - 2 ) ^{2}}{5} = 20

Multiply both sides by

5

10 (12 + 2 x) - (3 x - 2)^{2} = 100

Expand

10 (12 + 2 x) - (3 x - 2)^{2} : - 9 x^{2} + 32 x + 116

- 9 x^{2} + 32 x + 116 = 100

Move

100

to the left side

- 9 x^{2} + 32 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 9 ) \cdot 16}{2 ( - 9 )}

3 2^{2} - 4 (- 9) \cdot 16 = 40

x_{1, 2} = \frac{- 32 \pm 40}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- 32 + 40}{2 ( - 9 )}, x_{2} = \frac{- 32 - 40}{2 ( - 9 )}

x = \frac{- 32 + 40}{2 ( - 9 )} : - \frac{4}{9}

x = \frac{- 32 - 40}{2 ( - 9 )} : 4

The solutions to the quadratic equation are:

x = - \frac{4}{9}, x = 4

D, D^{2} y - 4 e^{5 x} = 0

4 \cdot 8^{2} = d^{2} + d^{2}

x, \frac{x ^{2} + y ^{3}}{y} - 4 x^{2} y = 4

x^{2} - 5 x + 1 = 0

x^{2} - 8 x + 15

What are the solutions to the equation 2(12+2x)-((3x-2)^2)/5 =20 ?
The solutions to the equation 2(12+2x)-((3x-2)^2)/5 =20 are x=-4/9 ,x=4
Find the zeros of 2(12+2x)-((3x-2)^2)/5 =20
The zeros of 2(12+2x)-((3x-2)^2)/5 =20 are x=-4/9 ,x=4