What are the solutions to the equation 12x+12=3x^2+3x ?

The solutions to the equation 12x+12=3x^2+3x are x=4,x=-1

Find the zeros of 12x+12=3x^2+3x

The zeros of 12x+12=3x^2+3x are x=4,x=-1

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

x = 4, x = - 1

Solution steps

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

Switch sides

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

Move

12

to the left side

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

Move

12 x

to the left side

3 x^{2} - 9 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 (- 12) = 15

x_{1, 2} = \frac{- ( - 9 ) \pm 15}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 15}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 15}{2 \cdot 3}

x = \frac{- ( - 9 ) + 15}{2 \cdot 3} : 4

x = \frac{- ( - 9 ) - 15}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

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

2 x^{2} - 7 x - 15 = 0

\frac{x ^{2}}{2} - \frac{3 x}{4} = 0

120000 = 350 x - 0.25 x^{2}

x^{2} + 6 x - 36 = 0

What are the solutions to the equation 12x+12=3x^2+3x ?
The solutions to the equation 12x+12=3x^2+3x are x=4,x=-1
Find the zeros of 12x+12=3x^2+3x
The zeros of 12x+12=3x^2+3x are x=4,x=-1