What are the solutions to the equation 0=2(x+5)^2-7 ?

The solutions to the equation 0=2(x+5)^2-7 are x=sqrt(7/2)-5,x=-sqrt(7/2)-5

Find the zeros of 0=2(x+5)^2-7

The zeros of 0=2(x+5)^2-7 are x=sqrt(7/2)-5,x=-sqrt(7/2)-5

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0 = 2 (x + 5)^{2} - 7

x = \frac{7}{2} - 5, x = - \frac{7}{2} - 5

Decimal Notation

x = - 3.12917 \dots, x = - 6.87082 \dots

Solution steps

0 = 2 (x + 5)^{2} - 7

Switch sides

2 (x + 5)^{2} - 7 = 0

Move

7

to the right side

2 (x + 5)^{2} = 7

Divide both sides by

2

(x + 5)^{2} = \frac{7}{2}

For

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

the solutions are

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

Solve

x + 5 = \frac{7}{2} : x = \frac{7}{2} - 5

Solve

x + 5 = - \frac{7}{2} : x = - \frac{7}{2} - 5

The solutions to the quadratic equation are:

x = \frac{7}{2} - 5, x = - \frac{7}{2} - 5

4 x^{2} - 15 x + 11 = 0

0 = 2 x^{2} + 3 x + 1

x^{2} - 14 x + 8 = 0

y, x = (y - 4)^{2}

3 a^{2} + 10 a + 8 = 0

What are the solutions to the equation 0=2(x+5)^2-7 ?
The solutions to the equation 0=2(x+5)^2-7 are x=sqrt(7/2)-5,x=-sqrt(7/2)-5
Find the zeros of 0=2(x+5)^2-7
The zeros of 0=2(x+5)^2-7 are x=sqrt(7/2)-5,x=-sqrt(7/2)-5