What are the solutions to the equation 2x^2=2x+3 ?

The solutions to the equation 2x^2=2x+3 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2

Find the zeros of 2x^2=2x+3

The zeros of 2x^2=2x+3 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2

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2 x^{2} = 2 x + 3

x = \frac{1 + 7}{2}, x = \frac{1 - 7}{2}

Decimal Notation

x = 1.82287 \dots, x = - 0.82287 \dots

Solution steps

2 x^{2} = 2 x + 3

Move

3

to the left side

2 x^{2} - 3 = 2 x

Move

2 x

to the left side

2 x^{2} - 3 - 2 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 2 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 (- 3) = 27

x_{1, 2} = \frac{- ( - 2 ) \pm 2 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 7}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 7}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 7}{2 \cdot 2} : \frac{1 + 7}{2}

x = \frac{- ( - 2 ) - 2 7}{2 \cdot 2} : \frac{1 - 7}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 7}{2}, x = \frac{1 - 7}{2}

2 - (z - 2)^{2} = z^{2}

v^{2} + 7 v + 6 = 0

a^{2} - 10 a - 24 = 0

3 d e + 6 df + 12 d g

\frac{( 4 n ) ^{2}}{- 4 n ^{- 5}}

What are the solutions to the equation 2x^2=2x+3 ?
The solutions to the equation 2x^2=2x+3 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2
Find the zeros of 2x^2=2x+3
The zeros of 2x^2=2x+3 are x=(1+sqrt(7))/2 ,x=(1-sqrt(7))/2