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

The solutions to the equation (3x-1)^2+x(x-3)=-1 are x= 1/2 ,x= 2/5

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

The zeros of (3x-1)^2+x(x-3)=-1 are x= 1/2 ,x= 2/5

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(3 x - 1)^{2} + x (x - 3) = - 1

x = \frac{1}{2}, x = \frac{2}{5}

Solution steps

(3 x - 1)^{2} + x (x - 3) = - 1

Expand

(3 x - 1)^{2} + x (x - 3) : 10 x^{2} - 9 x + 1

10 x^{2} - 9 x + 1 = - 1

Move

1

to the left side

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

Solve with the quadratic formula

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

(- 9)^{2} - 4 \cdot 10 \cdot 2 = 1

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

Separate the solutions

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

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

x = \frac{- ( - 9 ) - 1}{2 \cdot 10} : \frac{2}{5}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = \frac{2}{5}

5 t^{2} + 3 t - 2 = 0

25 x^{2} + 30 x + 12

t, h = v t - g t^{2}

9 x^{2} - 12 x - 14 = 0

x^{2} - 13 x + 1 = - 9 x + 6

What are the solutions to the equation (3x-1)^2+x(x-3)=-1 ?
The solutions to the equation (3x-1)^2+x(x-3)=-1 are x= 1/2 ,x= 2/5
Find the zeros of (3x-1)^2+x(x-3)=-1
The zeros of (3x-1)^2+x(x-3)=-1 are x= 1/2 ,x= 2/5