What are the solutions to the equation 4(5m+3)^2-28=0 ?

The solutions to the equation 4(5m+3)^2-28=0 are m=(sqrt(7)-3)/5 ,m=(-sqrt(7)-3)/5

Find the zeros of 4(5m+3)^2-28=0

The zeros of 4(5m+3)^2-28=0 are m=(sqrt(7)-3)/5 ,m=(-sqrt(7)-3)/5

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4 (5 m + 3)^{2} - 28 = 0

m = \frac{7 - 3}{5}, m = \frac{- 7 - 3}{5}

Decimal

m = - 0.07084 \dots, m = - 1.12915 \dots

Solution steps

4 (5 m + 3)^{2} - 28 = 0

Move

28

to the right side

4 (5 m + 3)^{2} = 28

Divide both sides by

4

(5 m + 3)^{2} = 7

For

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

the solutions are

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

Solve

5 m + 3 = 7 : m = \frac{7 - 3}{5}

Solve

5 m + 3 = - 7 : m = \frac{- 7 - 3}{5}

The solutions to the quadratic equation are:

m = \frac{7 - 3}{5}, m = \frac{- 7 - 3}{5}

C = 400 - 5 x + 0.125 x^{2}

(6 x - 1)^{2} = - 8

9^{2} = 4. 5^{2} + x^{2}

81 x^{2} - 256 = 0

x^{2} + 23 x + 102 = 0

What are the solutions to the equation 4(5m+3)^2-28=0 ?
The solutions to the equation 4(5m+3)^2-28=0 are m=(sqrt(7)-3)/5 ,m=(-sqrt(7)-3)/5
Find the zeros of 4(5m+3)^2-28=0
The zeros of 4(5m+3)^2-28=0 are m=(sqrt(7)-3)/5 ,m=(-sqrt(7)-3)/5