What are the solutions to the equation -2(x-3)^2+7=-4 ?

The solutions to the equation -2(x-3)^2+7=-4 are x=sqrt(11/2)+3,x=-sqrt(11/2)+3

Find the zeros of -2(x-3)^2+7=-4

The zeros of -2(x-3)^2+7=-4 are x=sqrt(11/2)+3,x=-sqrt(11/2)+3

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- 2 (x - 3)^{2} + 7 = - 4

x = \frac{11}{2} + 3, x = - \frac{11}{2} + 3

Decimal

x = 5.34520 \dots, x = 0.65479 \dots

Solution steps

- 2 (x - 3)^{2} + 7 = - 4

Move

7

to the right side

- 2 (x - 3)^{2} = - 11

Divide both sides by

- 2

(x - 3)^{2} = \frac{11}{2}

For

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

the solutions are

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

Solve

x - 3 = \frac{11}{2} : x = \frac{11}{2} + 3

Solve

x - 3 = - \frac{11}{2} : x = - \frac{11}{2} + 3

The solutions to the quadratic equation are:

x = \frac{11}{2} + 3, x = - \frac{11}{2} + 3

- 5 x^{2} + 3 0^{2} = 0

2 q^{2} + 10 q - 3 = 0

3 x^{2} + 10 x + 9 = - 3 x + 5

(x + 36)^{2} - 38 = 251

5 x^{2} + 2 x - 3

What are the solutions to the equation -2(x-3)^2+7=-4 ?
The solutions to the equation -2(x-3)^2+7=-4 are x=sqrt(11/2)+3,x=-sqrt(11/2)+3
Find the zeros of -2(x-3)^2+7=-4
The zeros of -2(x-3)^2+7=-4 are x=sqrt(11/2)+3,x=-sqrt(11/2)+3