What are the solutions to the equation -3x^2-9x+12=0 ?

The solutions to the equation -3x^2-9x+12=0 are x=-4,x=1

Find the zeros of -3x^2-9x+12=0

The zeros of -3x^2-9x+12=0 are x=-4,x=1

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

x = - 4, x = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 4, x = 1

(2 x + 1) (4 x - 3) = 0

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

- 2 x^{2} - 28 x + 30 = 0

(5 x^{2} + x)^{2} = (3 x + 3)^{2}

x, x^{2} + y^{2} - 6 x - 8 y = - 25

What are the solutions to the equation -3x^2-9x+12=0 ?
The solutions to the equation -3x^2-9x+12=0 are x=-4,x=1
Find the zeros of -3x^2-9x+12=0
The zeros of -3x^2-9x+12=0 are x=-4,x=1