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

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

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

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

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

x = 5, x = 1

Solution steps

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

Switch sides

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

Move

5

to the left side

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

Move

2 x

to the left side

x^{2} - 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 5, x = 1

e^{t} (e^{t})

(x + 3) (x - 3) = 16

\frac{2 x}{3} + 4 \leq 12 + 2 x

5.0 E 14 = 3.51 x^{2} + 24.4 x + 780

5 x^{2} - 10 x = 0

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