What are the solutions to the equation x^2-6x+5=-x^2+4x ?

The solutions to the equation x^2-6x+5=-x^2+4x are x=(5+sqrt(15))/2 ,x=(5-sqrt(15))/2

Find the zeros of x^2-6x+5=-x^2+4x

The zeros of x^2-6x+5=-x^2+4x are x=(5+sqrt(15))/2 ,x=(5-sqrt(15))/2

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x^{2} - 6 x + 5 = - x^{2} + 4 x

x = \frac{5 + 15}{2}, x = \frac{5 - 15}{2}

Decimal Notation

x = 4.43649 \dots, x = 0.56350 \dots

Solution steps

x^{2} - 6 x + 5 = - x^{2} + 4 x

Move

4 x

to the left side

x^{2} - 10 x + 5 = - x^{2}

Move

x^{2}

to the left side

2 x^{2} - 10 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 \cdot 5 = 215

x_{1, 2} = \frac{- ( - 10 ) \pm 2 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 15}{2 \cdot 2}, x_{2} = \frac{- ( - 10 ) - 2 15}{2 \cdot 2}

x = \frac{- ( - 10 ) + 2 15}{2 \cdot 2} : \frac{5 + 15}{2}

x = \frac{- ( - 10 ) - 2 15}{2 \cdot 2} : \frac{5 - 15}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 15}{2}, x = \frac{5 - 15}{2}

x, z = - x^{2} - 2 y

x^{2} + 5 x + 2 = - 4

3 x (x - 2) = - 7

x^{2} - 7 x - 3

3 v^{2} + 17 v + 17 = - 2

What are the solutions to the equation x^2-6x+5=-x^2+4x ?
The solutions to the equation x^2-6x+5=-x^2+4x are x=(5+sqrt(15))/2 ,x=(5-sqrt(15))/2
Find the zeros of x^2-6x+5=-x^2+4x
The zeros of x^2-6x+5=-x^2+4x are x=(5+sqrt(15))/2 ,x=(5-sqrt(15))/2