What are the solutions to the equation 2x+7=3x^2 ?

The solutions to the equation 2x+7=3x^2 are x=(1+sqrt(22))/3 ,x=(1-sqrt(22))/3

Find the zeros of 2x+7=3x^2

The zeros of 2x+7=3x^2 are x=(1+sqrt(22))/3 ,x=(1-sqrt(22))/3

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2 x + 7 = 3 x^{2}

x = \frac{1 + 22}{3}, x = \frac{1 - 22}{3}

Decimal Notation

x = 1.89680 \dots, x = - 1.23013 \dots

Solution steps

2 x + 7 = 3 x^{2}

Switch sides

3 x^{2} = 2 x + 7

Move

7

to the left side

3 x^{2} - 7 = 2 x

Move

2 x

to the left side

3 x^{2} - 7 - 2 x = 0

Write in the standard form

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

3 x^{2} - 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 7) = 222

x_{1, 2} = \frac{- ( - 2 ) \pm 2 22}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 22}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 22}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 22}{2 \cdot 3} : \frac{1 + 22}{3}

x = \frac{- ( - 2 ) - 2 22}{2 \cdot 3} : \frac{1 - 22}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 22}{3}, x = \frac{1 - 22}{3}

36 y^{2} - 24 y + 85 = 0

5 v^{2} - 20 v = 0

(n) = n^{2} - 4 n + 20

- x^{2} + 7 x = 11.25

w^{2} - 10 w

What are the solutions to the equation 2x+7=3x^2 ?
The solutions to the equation 2x+7=3x^2 are x=(1+sqrt(22))/3 ,x=(1-sqrt(22))/3
Find the zeros of 2x+7=3x^2
The zeros of 2x+7=3x^2 are x=(1+sqrt(22))/3 ,x=(1-sqrt(22))/3