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

The solutions to the equation (x+3)(x-4)=3x are x=6,x=-2

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

The zeros of (x+3)(x-4)=3x are x=6,x=-2

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(x + 3) (x - 4) = 3 x

x = 6, x = - 2

Solution steps

(x + 3) (x - 4) = 3 x

Expand

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

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

Move

3 x

to the left side

x^{2} - 4 x - 12 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 12) = 8

x_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 1}

x = \frac{- ( - 4 ) + 8}{2 \cdot 1} : 6

x = \frac{- ( - 4 ) - 8}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 6, x = - 2

x^{1 + l o g_{x} (64)} = 8 x^{2}

10 x^{2} + 34 x + 12 = 0

x, z = 49 - x^{2} - y^{2}

- 3 x^{2} + 8 = 2 x

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

What are the solutions to the equation (x+3)(x-4)=3x ?
The solutions to the equation (x+3)(x-4)=3x are x=6,x=-2
Find the zeros of (x+3)(x-4)=3x
The zeros of (x+3)(x-4)=3x are x=6,x=-2