What are the solutions to the equation 4(x+1)^2-31=45 ?

The solutions to the equation 4(x+1)^2-31=45 are x=sqrt(19)-1,x=-sqrt(19)-1

Find the zeros of 4(x+1)^2-31=45

The zeros of 4(x+1)^2-31=45 are x=sqrt(19)-1,x=-sqrt(19)-1

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4 (x + 1)^{2} - 31 = 45

x = 19 - 1, x = - 19 - 1

Decimal Notation

x = 3.35889 \dots, x = - 5.35889 \dots

Solution steps

4 (x + 1)^{2} - 31 = 45

Move

31

to the right side

4 (x + 1)^{2} = 76

Divide both sides by

4

(x + 1)^{2} = 19

For

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

the solutions are

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

Solve

x + 1 = 19 : x = 19 - 1

Solve

x + 1 = - 19 : x = - 19 - 1

The solutions to the quadratic equation are:

x = 19 - 1, x = - 19 - 1

4 x^{2} - 20 x = - 17

3 x^{2} - 21 = 3

64 x^{2} + 17 = 0

2 x^{2} + 5 x = 7

x^{2} + 0.7 = 2.4 x

What are the solutions to the equation 4(x+1)^2-31=45 ?
The solutions to the equation 4(x+1)^2-31=45 are x=sqrt(19)-1,x=-sqrt(19)-1
Find the zeros of 4(x+1)^2-31=45
The zeros of 4(x+1)^2-31=45 are x=sqrt(19)-1,x=-sqrt(19)-1