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

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

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

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

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x^{2} - 3 = \frac{( x - 3 )}{2}

x = \frac{3}{2}, x = - 1

Decimal Notation

x = 1.5, x = - 1

Solution steps

x^{2} - 3 = \frac{( x - 3 )}{2}

Multiply both sides by

2

2 x^{2} - 6 = x - 3

Move

3

to the left side

2 x^{2} - 3 = x

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 3) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 2}

x = \frac{- ( - 1 ) + 5}{2 \cdot 2} : \frac{3}{2}

x = \frac{- ( - 1 ) - 5}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = - 1

1 x^{2} + 20 x = 56

x^{2} - x - b = 0

x, 1^{2} + x^{2} = c \cdot p^{2}

x, 5 x^{2} - 7 x y - 3 y^{2} = 0

x, f (x + 1) = (x + 1)^{2} - 1

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