What are the solutions to the equation x+10x^2-12x=x^2+2x ?

The solutions to the equation x+10x^2-12x=x^2+2x are x= 13/9 ,x=0

Find the zeros of x+10x^2-12x=x^2+2x

The zeros of x+10x^2-12x=x^2+2x are x= 13/9 ,x=0

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x + 10 x^{2} - 12 x = x^{2} + 2 x

x = \frac{13}{9}, x = 0

Decimal Notation

x = 1.44444 \dots, x = 0

Solution steps

x + 10 x^{2} - 12 x = x^{2} + 2 x

Refine

10 x^{2} - 11 x = x^{2} + 2 x

Move

2 x

to the left side

10 x^{2} - 13 x = x^{2}

Move

x^{2}

to the left side

9 x^{2} - 13 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 9 \cdot 0}{2 \cdot 9}

(- 13)^{2} - 4 \cdot 9 \cdot 0 = 13

x_{1, 2} = \frac{- ( - 13 ) \pm 13}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 13}{2 \cdot 9}, x_{2} = \frac{- ( - 13 ) - 13}{2 \cdot 9}

x = \frac{- ( - 13 ) + 13}{2 \cdot 9} : \frac{13}{9}

x = \frac{- ( - 13 ) - 13}{2 \cdot 9} : 0

The solutions to the quadratic equation are:

x = \frac{13}{9}, x = 0

x^{2} - \frac{9}{2} x - \frac{24}{5} = 0

x, x^{2} + y^{2} - 4 x + 12 y + 40 = 0

r, (x - h)^{2} + (y - h)^{2} = r^{2}

4 x^{2} - 18 x + 18 = 0

6^{2} = x^{2} + 3^{2}

What are the solutions to the equation x+10x^2-12x=x^2+2x ?
The solutions to the equation x+10x^2-12x=x^2+2x are x= 13/9 ,x=0
Find the zeros of x+10x^2-12x=x^2+2x
The zeros of x+10x^2-12x=x^2+2x are x= 13/9 ,x=0