What are the solutions to the equation 3x^2-12x+7=-2 ?

The solutions to the equation 3x^2-12x+7=-2 are x=3,x=1

Find the zeros of 3x^2-12x+7=-2

The zeros of 3x^2-12x+7=-2 are x=3,x=1

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

x = 3, x = 1

Solution steps

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

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 3 \cdot 9 = 6

x_{1, 2} = \frac{- ( - 12 ) \pm 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 3}

x = \frac{- ( - 12 ) + 6}{2 \cdot 3} : 3

x = \frac{- ( - 12 ) - 6}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

0.5 = \frac{1}{2} \cdot 9.8 \cdot x^{2}

(5 x^{2} + 2 x - 11) - (- 4 x^{2} + 15 x + 1) = 0

10 x^{2} + 25 x = 0

x, F = (m - 2) x^{2} + (n - 1) x + (p - 3)

(x + 9)^{2} = 12

What are the solutions to the equation 3x^2-12x+7=-2 ?
The solutions to the equation 3x^2-12x+7=-2 are x=3,x=1
Find the zeros of 3x^2-12x+7=-2
The zeros of 3x^2-12x+7=-2 are x=3,x=1