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

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

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

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

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2 x^{2} - x = 3

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

Decimal Notation

x = 1.5, x = - 1

Solution steps

2 x^{2} - x = 3

Move

3

to the left side

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

\frac{n}{24} = - 2

2^{9} - 1

\frac{32}{9}

p - 5 (p + 4) = p - (8 - p)

\frac{n}{- 4} = - 34

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