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

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

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

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

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

x = - 1

Solution steps

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

Expand

2 x (x - 1) : 2 x^{2} - 2 x

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

Move

1

to the left side

2 x^{2} - 2 x - 1 = 3 x^{2}

Move

3 x^{2}

to the left side

- x^{2} - 2 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) ( - 1 )}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) (- 1) = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 ( - 1 )}

x = \frac{- ( - 2 )}{2 ( - 1 )}

\frac{- ( - 2 )}{2 ( - 1 )} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

π r^{2} h

- 15 x^{2} - 3 = 2 (7 x^{2} - 1.5)

y, x^{2} = 2 y^{2}

(x - 1)^{2} - 12 = 26

x^{2} - 4 x = 4 x

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