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

Switch sides

x^{2} - 4 x = 5

Move

5

to the left side

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

4 p^{2} + 19 p + 12 = - 5

8 y^{2} - 24 y = 0

x, f (x + 2) = x^{2} - 2 x + 5

y, x = y^{2} + 2 y - 1

x^{2} - 2 x + 6 = 2 x + 11

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