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

The solutions to the equation 4-x^2=5x are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2

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

The zeros of 4-x^2=5x are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2

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4 - x^{2} = 5 x

x = - \frac{5 + 41}{2}, x = \frac{41 - 5}{2}

Decimal Notation

x = - 5.70156 \dots, x = 0.70156 \dots

Solution steps

4 - x^{2} = 5 x

Move

5 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 (- 1) \cdot 4 = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 ( - 1 )}, x_{2} = \frac{- ( - 5 ) - 41}{2 ( - 1 )}

x = \frac{- ( - 5 ) + 41}{2 ( - 1 )} : - \frac{5 + 41}{2}

x = \frac{- ( - 5 ) - 41}{2 ( - 1 )} : \frac{41 - 5}{2}

The solutions to the quadratic equation are:

x = - \frac{5 + 41}{2}, x = \frac{41 - 5}{2}

(2 x - 3) 2 + x^{2} = (3 x + 1) (3 x - 1) - 6

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

3 x^{2} + 5 x - 138 = 0

q, p = \frac{q ^{2} + 20}{10}

x (x + 8) + 2900 = 0

What are the solutions to the equation 4-x^2=5x ?
The solutions to the equation 4-x^2=5x are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2
Find the zeros of 4-x^2=5x
The zeros of 4-x^2=5x are x=-(5+sqrt(41))/2 ,x=(sqrt(41)-5)/2