What are the solutions to the equation 25-x^2=4x ?

The solutions to the equation 25-x^2=4x are x=-2-sqrt(29),x=sqrt(29)-2

Find the zeros of 25-x^2=4x

The zeros of 25-x^2=4x are x=-2-sqrt(29),x=sqrt(29)-2

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25 - x^{2} = 4 x

x = - 2 - 29, x = 29 - 2

Decimal Notation

x = - 7.38516 \dots, x = 3.38516 \dots

Solution steps

25 - x^{2} = 4 x

Move

4 x

to the left side

25 - x^{2} - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} - 4 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) \cdot 25}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) \cdot 25 = 229

x_{1, 2} = \frac{- ( - 4 ) \pm 2 29}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 29}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 2 29}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 2 29}{2 ( - 1 )} : - 2 - 29

x = \frac{- ( - 4 ) - 2 29}{2 ( - 1 )} : 29 - 2

The solutions to the quadratic equation are:

x = - 2 - 29, x = 29 - 2

C, E = M C^{2}

(y - 2)^{2} - 10 (y - 2) = - 9

5 n (n - 4) = 0

12 x^{2} + 23 x + 10 = 0

x^{2} + x - 32 = 0

What are the solutions to the equation 25-x^2=4x ?
The solutions to the equation 25-x^2=4x are x=-2-sqrt(29),x=sqrt(29)-2
Find the zeros of 25-x^2=4x
The zeros of 25-x^2=4x are x=-2-sqrt(29),x=sqrt(29)-2