What are the solutions to the equation x^2+4x=x+10 ?

The solutions to the equation x^2+4x=x+10 are x=2,x=-5

Find the zeros of x^2+4x=x+10

The zeros of x^2+4x=x+10 are x=2,x=-5

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x^{2} + 4 x = x + 10

x = 2, x = - 5

Solution steps

x^{2} + 4 x = x + 10

Move

10

to the left side

x^{2} + 4 x - 10 = x

Move

x

to the left side

x^{2} + 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 1}, x_{2} = \frac{- 3 - 7}{2 \cdot 1}

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

x = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

x^{2} + 5 x + 9 = 0

x^{2} + 6 x + 13

x^{2} + 17 = 0

4 x^{2} + 9 = 105

x^{2} + 3 x = - 5

What are the solutions to the equation x^2+4x=x+10 ?
The solutions to the equation x^2+4x=x+10 are x=2,x=-5
Find the zeros of x^2+4x=x+10
The zeros of x^2+4x=x+10 are x=2,x=-5