What are the solutions to the equation 23/2 =3x^2-x ?

The solutions to the equation 23/2 =3x^2-x are x=(1+sqrt(139))/6 ,x=(1-sqrt(139))/6

Find the zeros of 23/2 =3x^2-x

The zeros of 23/2 =3x^2-x are x=(1+sqrt(139))/6 ,x=(1-sqrt(139))/6

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\frac{23}{2} = 3 x^{2} - x

x = \frac{1 + 139}{6}, x = \frac{1 - 139}{6}

Decimal Notation

x = 2.13163 \dots, x = - 1.79830 \dots

Solution steps

\frac{23}{2} = 3 x^{2} - x

Multiply both sides by

2

23 = 6 x^{2} - 2 x

Switch sides

6 x^{2} - 2 x = 23

Move

23

to the left side

6 x^{2} - 2 x - 23 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 6 ( - 23 )}{2 \cdot 6}

(- 2)^{2} - 4 \cdot 6 (- 23) = 2139

x_{1, 2} = \frac{- ( - 2 ) \pm 2 139}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 139}{2 \cdot 6}, x_{2} = \frac{- ( - 2 ) - 2 139}{2 \cdot 6}

x = \frac{- ( - 2 ) + 2 139}{2 \cdot 6} : \frac{1 + 139}{6}

x = \frac{- ( - 2 ) - 2 139}{2 \cdot 6} : \frac{1 - 139}{6}

The solutions to the quadratic equation are:

x = \frac{1 + 139}{6}, x = \frac{1 - 139}{6}

\frac{4}{3} \cdot π \cdot 1 0^{3}

\frac{1}{3} - 3

\frac{- 7 ( - 10 ) ^{2} - 5 ( - 10 ) - 22}{- 10 + 1}

\frac{30 + 20 i}{1 + 5 i}

y + (4 (35) + 5) + (2 y - 55) = 180

What are the solutions to the equation 23/2 =3x^2-x ?
The solutions to the equation 23/2 =3x^2-x are x=(1+sqrt(139))/6 ,x=(1-sqrt(139))/6
Find the zeros of 23/2 =3x^2-x
The zeros of 23/2 =3x^2-x are x=(1+sqrt(139))/6 ,x=(1-sqrt(139))/6