What are the solutions to the equation 6x^2+9x=12 ?

The solutions to the equation 6x^2+9x=12 are x=(-3+sqrt(41))/4 ,x=-(3+sqrt(41))/4

Find the zeros of 6x^2+9x=12

The zeros of 6x^2+9x=12 are x=(-3+sqrt(41))/4 ,x=-(3+sqrt(41))/4

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6 x^{2} + 9 x = 12

x = \frac{- 3 + 41}{4}, x = - \frac{3 + 41}{4}

Decimal Notation

x = 0.85078 \dots, x = - 2.35078 \dots

Solution steps

6 x^{2} + 9 x = 12

Move

12

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 6 ( - 12 )}{2 \cdot 6}

9^{2} - 4 \cdot 6 (- 12) = 341

x_{1, 2} = \frac{- 9 \pm 3 41}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 9 + 3 41}{2 \cdot 6}, x_{2} = \frac{- 9 - 3 41}{2 \cdot 6}

x = \frac{- 9 + 3 41}{2 \cdot 6} : \frac{- 3 + 41}{4}

x = \frac{- 9 - 3 41}{2 \cdot 6} : - \frac{3 + 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 41}{4}, x = - \frac{3 + 41}{4}

(y - 5)^{2} = 49

x^{2} - 30 x + 150 = 0

24 x^{2} - 336 x + 864 = 0

0 = 3 x^{2} - 20 x + 25

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

What are the solutions to the equation 6x^2+9x=12 ?
The solutions to the equation 6x^2+9x=12 are x=(-3+sqrt(41))/4 ,x=-(3+sqrt(41))/4
Find the zeros of 6x^2+9x=12
The zeros of 6x^2+9x=12 are x=(-3+sqrt(41))/4 ,x=-(3+sqrt(41))/4