What are the solutions to the equation x^2-3(x+4)=-2x ?

The solutions to the equation x^2-3(x+4)=-2x are x=4,x=-3

Find the zeros of x^2-3(x+4)=-2x

The zeros of x^2-3(x+4)=-2x are x=4,x=-3

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x^{2} - 3 (x + 4) = - 2 x

x = 4, x = - 3

Solution steps

x^{2} - 3 (x + 4) = - 2 x

Expand

x^{2} - 3 (x + 4) : x^{2} - 3 x - 12

x^{2} - 3 x - 12 = - 2 x

Move

2 x

to the left side

x^{2} - x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = - 3

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

x^{2} - 18 x - 7 = 0

15 x - 10 = 6 x (x - 2) + (- x + 3)

2 y^{2} + 5 y = 0

x, f^{- 1} = x^{2} + 7 x + 10

What are the solutions to the equation x^2-3(x+4)=-2x ?
The solutions to the equation x^2-3(x+4)=-2x are x=4,x=-3
Find the zeros of x^2-3(x+4)=-2x
The zeros of x^2-3(x+4)=-2x are x=4,x=-3