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

The solutions to the equation 2x^2+6=9x are x=(9+sqrt(33))/4 ,x=(9-sqrt(33))/4

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

The zeros of 2x^2+6=9x are x=(9+sqrt(33))/4 ,x=(9-sqrt(33))/4

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2 x^{2} + 6 = 9 x

x = \frac{9 + 33}{4}, x = \frac{9 - 33}{4}

Decimal Notation

x = 3.68614 \dots, x = 0.81385 \dots

Solution steps

2 x^{2} + 6 = 9 x

Move

9 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 9)^{2} - 4 \cdot 2 \cdot 6 = 33

x_{1, 2} = \frac{- ( - 9 ) \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 33}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 33}{2 \cdot 2}

x = \frac{- ( - 9 ) + 33}{2 \cdot 2} : \frac{9 + 33}{4}

x = \frac{- ( - 9 ) - 33}{2 \cdot 2} : \frac{9 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 33}{4}, x = \frac{9 - 33}{4}

36 - 9

- 3 x \geq - 15 and 4 x + 3 \geq 15

∣ x + 2 ∣ + 5 \geq 9

\frac{7}{2} x^{2} + 7 x + 14 \geq 0

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

What are the solutions to the equation 2x^2+6=9x ?
The solutions to the equation 2x^2+6=9x are x=(9+sqrt(33))/4 ,x=(9-sqrt(33))/4
Find the zeros of 2x^2+6=9x
The zeros of 2x^2+6=9x are x=(9+sqrt(33))/4 ,x=(9-sqrt(33))/4