What are the solutions to the equation (20)(8)-(20-2x)(8-2x)=40 ?

The solutions to the equation (20)(8)-(20-2x)(8-2x)=40 are x=7-sqrt(39),x=7+sqrt(39)

Find the zeros of (20)(8)-(20-2x)(8-2x)=40

The zeros of (20)(8)-(20-2x)(8-2x)=40 are x=7-sqrt(39),x=7+sqrt(39)

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(20) (8) - (20 - 2 x) (8 - 2 x) = 40

x = 7 - 39, x = 7 + 39

Decimal Notation

x = 0.75500 \dots, x = 13.24499 \dots

Solution steps

(20) (8) - (20 - 2 x) (8 - 2 x) = 40

Expand

(20) (8) - (20 - 2 x) (8 - 2 x) : - 4 x^{2} + 56 x

- 4 x^{2} + 56 x = 40

Move

40

to the left side

- 4 x^{2} + 56 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 56 \pm 5 6 ^{2} - 4 ( - 4 ) ( - 40 )}{2 ( - 4 )}

5 6^{2} - 4 (- 4) (- 40) = 839

x_{1, 2} = \frac{- 56 \pm 8 39}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 56 + 8 39}{2 ( - 4 )}, x_{2} = \frac{- 56 - 8 39}{2 ( - 4 )}

x = \frac{- 56 + 8 39}{2 ( - 4 )} : 7 - 39

x = \frac{- 56 - 8 39}{2 ( - 4 )} : 7 + 39

The solutions to the quadratic equation are:

x = 7 - 39, x = 7 + 39

(3 s - 2) - 9 > - 5

- 9.1 < d - 6.3

(\frac{1}{16})^{- 4}

(\frac{9}{17})^{6}

\frac{x ^{3} + x ^{2} y + 3 y ^{3}}{x ^{3} + 3 x y ^{2}}

What are the solutions to the equation (20)(8)-(20-2x)(8-2x)=40 ?
The solutions to the equation (20)(8)-(20-2x)(8-2x)=40 are x=7-sqrt(39),x=7+sqrt(39)
Find the zeros of (20)(8)-(20-2x)(8-2x)=40
The zeros of (20)(8)-(20-2x)(8-2x)=40 are x=7-sqrt(39),x=7+sqrt(39)