What are the solutions to the equation (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) ?

The solutions to the equation (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) are x=sqrt(7),x=-sqrt(7)

Find the zeros of (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7)

The zeros of (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) are x=sqrt(7),x=-sqrt(7)

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\frac{49 + 25 - 9}{2 \cdot 7 \cdot 5} = \frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7}

x = 7, x = - 7

Decimal Notation

x = 2.64575 \dots, x = - 2.64575 \dots

Solution steps

\frac{49 + 25 - 9}{2 \cdot 7 \cdot 5} = \frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7}

Switch sides

\frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7} = \frac{49 + 25 - 9}{2 \cdot 7 \cdot 5}

Add/Subtract the numbers:

49 + 25 - 9 = 65

\frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7} = \frac{65}{2 \cdot 7 \cdot 5}

Multiply the numbers:

2 \cdot 7 \cdot 5 = 70

\frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7} = \frac{65}{70}

Cancel the common factor:

5

\frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7} = \frac{13}{14}

Multiply both sides by

2 \cdot 7 \cdot 7

- x^{2} + 98 = 91

Move

98

to the right side

- x^{2} = - 7

Divide both sides by

- 1

x^{2} = 7

For

x^{2} = f (a)

the solutions are

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

x = 7, x = - 7

x^{2} - 6 x - 4 = 0

e^{2} = 5

x, 0 = x^{2} + 5 x + 8

- x^{2} = - 49

16 + 9 = x^{2}

What are the solutions to the equation (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) ?
The solutions to the equation (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) are x=sqrt(7),x=-sqrt(7)
Find the zeros of (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7)
The zeros of (49+25-9)/(2*7*5)=(49+49-x^2)/(2*7*7) are x=sqrt(7),x=-sqrt(7)