What are the solutions to the equation 2(x+1)=3(x^2-1) ?

The solutions to the equation 2(x+1)=3(x^2-1) are x=-1,x= 5/3

Find the zeros of 2(x+1)=3(x^2-1)

The zeros of 2(x+1)=3(x^2-1) are x=-1,x= 5/3

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2 (x + 1) = 3 (x^{2} - 1)

x = - 1, x = \frac{5}{3}

Decimal Notation

x = - 1, x = 1.66666 \dots

Solution steps

2 (x + 1) = 3 (x^{2} - 1)

Move

3 (x^{2} - 1)

to the left side

2 (x + 1) - 3 (x^{2} - 1) = 0

Factor

2 (x + 1) - 3 (x^{2} - 1) : - (x + 1) (3 x - 5)

- (x + 1) (3 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 3 x - 5 = 0

Solve

x + 1 = 0 : x = - 1

Solve

3 x - 5 = 0 : x = \frac{5}{3}

The solutions to the quadratic equation are:

x = - 1, x = \frac{5}{3}

x^{2} - 56 = 7 x

100 m^{2} n^{6} = (x)^{2}

x^{2} + 10 x - 200 = 0

4 r^{2} + 4 r - 1 = 0

(x + 7) (x - 3) = 0

What are the solutions to the equation 2(x+1)=3(x^2-1) ?
The solutions to the equation 2(x+1)=3(x^2-1) are x=-1,x= 5/3
Find the zeros of 2(x+1)=3(x^2-1)
The zeros of 2(x+1)=3(x^2-1) are x=-1,x= 5/3