What are the solutions to the equation (0.2+x)(0.1+x)=x ?

The solutions to the equation (0.2+x)(0.1+x)=x are x=(0.7+sqrt(0.41))/2 ,x=(0.7-sqrt(0.41))/2

Find the zeros of (0.2+x)(0.1+x)=x

The zeros of (0.2+x)(0.1+x)=x are x=(0.7+sqrt(0.41))/2 ,x=(0.7-sqrt(0.41))/2

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(0.2 + x) (0.1 + x) = x

x = \frac{0.7 + 0.41}{2}, x = \frac{0.7 - 0.41}{2}

Solution steps

(0.2 + x) (0.1 + x) = x

Expand

(0.2 + x) (0.1 + x) : 0.02 + 0.3 x + x^{2}

0.02 + 0.3 x + x^{2} = x

Move

x

to the left side

x^{2} - 0.7 x + 0.02 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 0.7 ) \pm ( - 0.7 ) ^{2} - 4 \cdot 1 \cdot 0.02}{2 \cdot 1}

(- 0.7)^{2} - 4 \cdot 1 \cdot 0.02 = 0.41

x_{1, 2} = \frac{- ( - 0.7 ) \pm 0.41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 0.7 ) + 0.41}{2 \cdot 1}, x_{2} = \frac{- ( - 0.7 ) - 0.41}{2 \cdot 1}

x = \frac{- ( - 0.7 ) + 0.41}{2 \cdot 1} : \frac{0.7 + 0.41}{2}

x = \frac{- ( - 0.7 ) - 0.41}{2 \cdot 1} : \frac{0.7 - 0.41}{2}

The solutions to the quadratic equation are:

x = \frac{0.7 + 0.41}{2}, x = \frac{0.7 - 0.41}{2}

(x^{- \frac{1}{3}})^{2}

\frac{1}{2} \cdot \frac{1}{6} \cdot \frac{1}{2}

\frac{10 x - 5}{2 x - 4} < 0

x^{2} + x + x y + y

2 x^{2} - 8 x

What are the solutions to the equation (0.2+x)(0.1+x)=x ?
The solutions to the equation (0.2+x)(0.1+x)=x are x=(0.7+sqrt(0.41))/2 ,x=(0.7-sqrt(0.41))/2
Find the zeros of (0.2+x)(0.1+x)=x
The zeros of (0.2+x)(0.1+x)=x are x=(0.7+sqrt(0.41))/2 ,x=(0.7-sqrt(0.41))/2