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

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

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

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

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x^{2} - 5 = 4 x

x = 5, x = - 1

Solution steps

x^{2} - 5 = 4 x

Move

4 x

to the left side

x^{2} - 5 - 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 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 5, x = - 1

(\frac{3 \cdot 8.314}{2}) (745 - 298)

14 x + 4 = 11 x + 22

9 x^{2} + 36 x = 0

x^{2} - x > 2

(3 c - 5)^{2} - 16 c^{2}

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