What are the solutions to the equation 2(3x^2-2x+2)=15*x^2 ?

The solutions to the equation 2(3x^2-2x+2)=15*x^2 are x=-(2(1+sqrt(10)))/9 ,x=(2(sqrt(10)-1))/9

Find the zeros of 2(3x^2-2x+2)=15*x^2

The zeros of 2(3x^2-2x+2)=15*x^2 are x=-(2(1+sqrt(10)))/9 ,x=(2(sqrt(10)-1))/9

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2 (3 x^{2} - 2 x + 2) = 15 \cdot x^{2}

x = - \frac{2 ( 1 + 10 )}{9}, x = \frac{2 ( 10 - 1 )}{9}

Solution steps

2 (3 x^{2} - 2 x + 2) = 15 x^{2}

Expand

2 (3 x^{2} - 2 x + 2) : 6 x^{2} - 4 x + 4

6 x^{2} - 4 x + 4 = 15 x^{2}

Move

15 \cdot x^{2}

to the left side

- 9 x^{2} - 4 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 9 ) \cdot 4}{2 ( - 9 )}

(- 4)^{2} - 4 (- 9) \cdot 4 = 410

x_{1, 2} = \frac{- ( - 4 ) \pm 4 10}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 10}{2 ( - 9 )}, x_{2} = \frac{- ( - 4 ) - 4 10}{2 ( - 9 )}

x = \frac{- ( - 4 ) + 4 10}{2 ( - 9 )} : - \frac{2 ( 1 + 10 )}{9}

x = \frac{- ( - 4 ) - 4 10}{2 ( - 9 )} : \frac{2 ( 10 - 1 )}{9}

The solutions to the quadratic equation are:

x = - \frac{2 ( 1 + 10 )}{9}, x = \frac{2 ( 10 - 1 )}{9}

5 n^{2} = - 8 n - 8

3 z^{2} - 27 = 0

3 x^{2} - 2 (a + b) x - (a + b)^{2} = 0

\frac{( y - 2 ) ^{2}}{3} = 8 - y

4 x^{2} + 4 x - 80 = 0

What are the solutions to the equation 2(3x^2-2x+2)=15*x^2 ?
The solutions to the equation 2(3x^2-2x+2)=15*x^2 are x=-(2(1+sqrt(10)))/9 ,x=(2(sqrt(10)-1))/9
Find the zeros of 2(3x^2-2x+2)=15*x^2
The zeros of 2(3x^2-2x+2)=15*x^2 are x=-(2(1+sqrt(10)))/9 ,x=(2(sqrt(10)-1))/9