What are the solutions to the equation -18-3x+6x^2=0 ?

The solutions to the equation -18-3x+6x^2=0 are x=2,x=-3/2

Find the zeros of -18-3x+6x^2=0

The zeros of -18-3x+6x^2=0 are x=2,x=-3/2

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- 18 - 3 x + 6 x^{2} = 0

x = 2, x = - \frac{3}{2}

Decimal Notation

x = 2, x = - 1.5

Solution steps

- 18 - 3 x + 6 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

6 x^{2} - 3 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 ( - 18 )}{2 \cdot 6}

(- 3)^{2} - 4 \cdot 6 (- 18) = 21

x_{1, 2} = \frac{- ( - 3 ) \pm 21}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 21}{2 \cdot 6}, x_{2} = \frac{- ( - 3 ) - 21}{2 \cdot 6}

x = \frac{- ( - 3 ) + 21}{2 \cdot 6} : 2

x = \frac{- ( - 3 ) - 21}{2 \cdot 6} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = 2, x = - \frac{3}{2}

3 (x - 2)^{2} = 6

x (2 x - 1) = 10

8 x^{2} - 22 x = 21

(x + 7) (x - 7) = 3 x

10 = 6 t - t^{2}

What are the solutions to the equation -18-3x+6x^2=0 ?
The solutions to the equation -18-3x+6x^2=0 are x=2,x=-3/2
Find the zeros of -18-3x+6x^2=0
The zeros of -18-3x+6x^2=0 are x=2,x=-3/2