What are the solutions to the equation x^2-4x=7x-28 ?

The solutions to the equation x^2-4x=7x-28 are x=7,x=4

Find the zeros of x^2-4x=7x-28

The zeros of x^2-4x=7x-28 are x=7,x=4

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x^{2} - 4 x = 7 x - 28

x = 7, x = 4

Solution steps

x^{2} - 4 x = 7 x - 28

Move

28

to the left side

x^{2} - 4 x + 28 = 7 x

Move

7 x

to the left side

x^{2} - 11 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

x_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

x = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

x = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 7, x = 4

- 2 r^{2} + 5 r - 22 = - 4 r^{2} + 6 r + 6

35 = \frac{n ( n - 3 )}{2}

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

- 0.5 x^{2} + 620 = 30 x + 270

4 x^{2} + 14 x - 3 = 3 x

What are the solutions to the equation x^2-4x=7x-28 ?
The solutions to the equation x^2-4x=7x-28 are x=7,x=4
Find the zeros of x^2-4x=7x-28
The zeros of x^2-4x=7x-28 are x=7,x=4