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

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

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

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

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3 x^{2} = 9 x + 12

x = 4, x = - 1

Solution steps

3 x^{2} = 9 x + 12

Move

12

to the left side

3 x^{2} - 12 = 9 x

Move

9 x

to the left side

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

Write in the standard form

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

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

13 = ∣ 2 - 3 x ∣

2 (5 w + 9) \leq 7 w + 3

∣ x + 6 ∣ > 10

(\frac{2}{3})^{4} (\frac{9}{4})^{4} (\frac{4}{3})^{4}

2 (x - 2) = - (4 - x)

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