What are the solutions to the equation x^2-x=3(7+x) ?

The solutions to the equation x^2-x=3(7+x) are x=7,x=-3

Find the zeros of x^2-x=3(7+x)

The zeros of x^2-x=3(7+x) are x=7,x=-3

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x^{2} - x = 3 (7 + x)

x = 7, x = - 3

Solution steps

x^{2} - x = 3 (7 + x)

Expand

3 (7 + x) : 21 + 3 x

x^{2} - x = 21 + 3 x

Move

3 x

to the left side

x^{2} - 4 x = 21

Move

21

to the left side

x^{2} - 4 x - 21 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

x_{1, 2} = \frac{- ( - 4 ) \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

x = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

x = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 7, x = - 3

y, f = x^{3} y^{2} - 2 x^{2} y + 3 x

4 a^{2} - 13 a + 3 = 0

- 13 - 5 x = - 5 x^{2}

y^{2} + 13 y + 42 = 0

x, 7 x^{2} + 3 x - 2 = 0

What are the solutions to the equation x^2-x=3(7+x) ?
The solutions to the equation x^2-x=3(7+x) are x=7,x=-3
Find the zeros of x^2-x=3(7+x)
The zeros of x^2-x=3(7+x) are x=7,x=-3