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

The solutions to the equation 2(x+2)^2+5=12 are x=sqrt(7/2)-2,x=-sqrt(7/2)-2

Find the zeros of 2(x+2)^2+5=12

The zeros of 2(x+2)^2+5=12 are x=sqrt(7/2)-2,x=-sqrt(7/2)-2

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2 (x + 2)^{2} + 5 = 12

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

Decimal Notation

x = - 0.12917 \dots, x = - 3.87082 \dots

Solution steps

2 (x + 2)^{2} + 5 = 12

Move

5

to the right side

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

Divide both sides by

2

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

2 n^{2} - 7 n - 21 = 0

(2 m - 1) x^{2} + m x + 15 = 0

2 - x (x + 4) = - 1 (x - 2)

(x + 7)^{2} = 15

1268.4 = 0.68 \cdot 100 \cdot X \cdot 1.79 \cdot (7.6 - 0.4 \cdot X)

What are the solutions to the equation 2(x+2)^2+5=12 ?
The solutions to the equation 2(x+2)^2+5=12 are x=sqrt(7/2)-2,x=-sqrt(7/2)-2
Find the zeros of 2(x+2)^2+5=12
The zeros of 2(x+2)^2+5=12 are x=sqrt(7/2)-2,x=-sqrt(7/2)-2