What are the solutions to the equation -2x^2-x+12=-3x^2+6x ?

The solutions to the equation -2x^2-x+12=-3x^2+6x are x=4,x=3

Find the zeros of -2x^2-x+12=-3x^2+6x

The zeros of -2x^2-x+12=-3x^2+6x are x=4,x=3

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- 2 x^{2} - x + 12 = - 3 x^{2} + 6 x

x = 4, x = 3

Solution steps

- 2 x^{2} - x + 12 = - 3 x^{2} + 6 x

Move

6 x

to the left side

- 2 x^{2} - 7 x + 12 = - 3 x^{2}

Move

3 x^{2}

to the left side

x^{2} - 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 3

x^{2} - 32 x + 240 = 0

\frac{1}{3} (x - 4)^{2} = - 3

2 x^{2} - 10 x + 20 = 9

x^{2} + 4 x = 16

(x + 5)^{2} + (x)^{2} = 1 9^{2}

What are the solutions to the equation -2x^2-x+12=-3x^2+6x ?
The solutions to the equation -2x^2-x+12=-3x^2+6x are x=4,x=3
Find the zeros of -2x^2-x+12=-3x^2+6x
The zeros of -2x^2-x+12=-3x^2+6x are x=4,x=3