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

The solutions to the equation 2x^2+9x=12 are x=(-9+sqrt(177))/4 ,x=(-9-sqrt(177))/4

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

The zeros of 2x^2+9x=12 are x=(-9+sqrt(177))/4 ,x=(-9-sqrt(177))/4

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2 x^{2} + 9 x = 12

x = \frac{- 9 + 177}{4}, x = \frac{- 9 - 177}{4}

Decimal Notation

x = 1.07603 \dots, x = - 5.57603 \dots

Solution steps

2 x^{2} + 9 x = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

9^{2} - 4 \cdot 2 (- 12) = 177

x_{1, 2} = \frac{- 9 \pm 177}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 9 + 177}{2 \cdot 2}, x_{2} = \frac{- 9 - 177}{2 \cdot 2}

x = \frac{- 9 + 177}{2 \cdot 2} : \frac{- 9 + 177}{4}

x = \frac{- 9 - 177}{2 \cdot 2} : \frac{- 9 - 177}{4}

The solutions to the quadratic equation are:

x = \frac{- 9 + 177}{4}, x = \frac{- 9 - 177}{4}

\frac{7 ^{2}}{2 5 ^{2}} + \frac{y ^{2}}{2 4 ^{2}} = 1

3 x^{2} + 80 x - 1600 = 0

4 (10 - x)^{2} = 5 x^{2}

x, 4 x^{2} - y^{2} + 4 z^{2} - 4 = 0

(5 x + 3) (2 x - 8) = 0

What are the solutions to the equation 2x^2+9x=12 ?
The solutions to the equation 2x^2+9x=12 are x=(-9+sqrt(177))/4 ,x=(-9-sqrt(177))/4
Find the zeros of 2x^2+9x=12
The zeros of 2x^2+9x=12 are x=(-9+sqrt(177))/4 ,x=(-9-sqrt(177))/4