What are the solutions to the equation 4(r-3)^2-6=0 ?

The solutions to the equation 4(r-3)^2-6=0 are r=sqrt(3/2)+3,r=-sqrt(3/2)+3

Find the zeros of 4(r-3)^2-6=0

The zeros of 4(r-3)^2-6=0 are r=sqrt(3/2)+3,r=-sqrt(3/2)+3

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4 (r - 3)^{2} - 6 = 0

r = \frac{3}{2} + 3, r = - \frac{3}{2} + 3

Decimal Notation

r = 4.22474 \dots, r = 1.77525 \dots

Solution steps

4 (r - 3)^{2} - 6 = 0

Move

6

to the right side

4 (r - 3)^{2} = 6

Divide both sides by

4

(r - 3)^{2} = \frac{3}{2}

For

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

the solutions are

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

Solve

r - 3 = \frac{3}{2} : r = \frac{3}{2} + 3

Solve

r - 3 = - \frac{3}{2} : r = - \frac{3}{2} + 3

The solutions to the quadratic equation are:

r = \frac{3}{2} + 3, r = - \frac{3}{2} + 3

x^{2} - (1 \cdot 1 0^{- 7}) x - (1 \cdot 1 0^{- 14}) = 0

(\frac{3 x ^{2}}{5})^{2}

- 4 \leq 4 x + 8

t^{2} + 4 t + 4

m^{2} + 10 m y + 25 y^{2}

What are the solutions to the equation 4(r-3)^2-6=0 ?
The solutions to the equation 4(r-3)^2-6=0 are r=sqrt(3/2)+3,r=-sqrt(3/2)+3
Find the zeros of 4(r-3)^2-6=0
The zeros of 4(r-3)^2-6=0 are r=sqrt(3/2)+3,r=-sqrt(3/2)+3