What are the solutions to the equation x^2+x+13=4+7x ?

The solutions to the equation x^2+x+13=4+7x are x=3

Find the zeros of x^2+x+13=4+7x

The zeros of x^2+x+13=4+7x are x=3

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x^{2} + x + 13 = 4 + 7 x

x = 3

Solution steps

x^{2} + x + 13 = 4 + 7 x

Move

7 x

to the left side

x^{2} - 6 x + 13 = 4

Move

4

to the left side

x^{2} - 6 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 9 = 0

x_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 6 )}{2 \cdot 1} = 3

x = 3

The solution to the quadratic equation is:

x = 3

12 y^{2} = - 5 y^{2} + 2

(x - 4) (x - 6) = - 1

3 x^{2} + 9 x

x, 10 x^{2} - 8 y - 8 y^{3} = 0

7 x + 1 = (2 x - 1)^{2}

What are the solutions to the equation x^2+x+13=4+7x ?
The solutions to the equation x^2+x+13=4+7x are x=3
Find the zeros of x^2+x+13=4+7x
The zeros of x^2+x+13=4+7x are x=3