What are the solutions to the equation 3x^2-18=2x+1 ?

The solutions to the equation 3x^2-18=2x+1 are x=(1+sqrt(58))/3 ,x=(1-sqrt(58))/3

Find the zeros of 3x^2-18=2x+1

The zeros of 3x^2-18=2x+1 are x=(1+sqrt(58))/3 ,x=(1-sqrt(58))/3

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3 x^{2} - 18 = 2 x + 1

x = \frac{1 + 58}{3}, x = \frac{1 - 58}{3}

Decimal Notation

x = 2.87192 \dots, x = - 2.20525 \dots

Solution steps

3 x^{2} - 18 = 2 x + 1

Move

1

to the left side

3 x^{2} - 19 = 2 x

Move

2 x

to the left side

3 x^{2} - 19 - 2 x = 0

Write in the standard form

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

3 x^{2} - 2 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 19 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 19) = 258

x_{1, 2} = \frac{- ( - 2 ) \pm 2 58}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 58}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 58}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 58}{2 \cdot 3} : \frac{1 + 58}{3}

x = \frac{- ( - 2 ) - 2 58}{2 \cdot 3} : \frac{1 - 58}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 58}{3}, x = \frac{1 - 58}{3}

m^{2} + 12 m + 20 = 0

- 3 x^{2} - 120 = 0

4 x^{2} + 3 x - 22 = 0

E, P = \frac{E ^{2}}{R}

6 + 2 x = 5 x^{2}

What are the solutions to the equation 3x^2-18=2x+1 ?
The solutions to the equation 3x^2-18=2x+1 are x=(1+sqrt(58))/3 ,x=(1-sqrt(58))/3
Find the zeros of 3x^2-18=2x+1
The zeros of 3x^2-18=2x+1 are x=(1+sqrt(58))/3 ,x=(1-sqrt(58))/3