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

The solutions to the equation 5-x^2=4x are x=-5,x=1

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

The zeros of 5-x^2=4x are x=-5,x=1

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5 - x^{2} = 4 x

x = - 5, x = 1

Solution steps

5 - x^{2} = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

- x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 5 = 6

x_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 6}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 6}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 6}{2 ( - 1 )} : - 5

x = \frac{- ( - 4 ) - 6}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 5, x = 1

a^{2} + 2 a - 3 = 0

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

28 x^{2} - 85 x + 63 = 0

v^{2} = - 256

k, (n - k) (- k) = k (n - k)

What are the solutions to the equation 5-x^2=4x ?
The solutions to the equation 5-x^2=4x are x=-5,x=1
Find the zeros of 5-x^2=4x
The zeros of 5-x^2=4x are x=-5,x=1