What are the solutions to the equation 2x^2+9x=78-3x ?

The solutions to the equation 2x^2+9x=78-3x are x=-3+4sqrt(3),x=-3-4sqrt(3)

Find the zeros of 2x^2+9x=78-3x

The zeros of 2x^2+9x=78-3x are x=-3+4sqrt(3),x=-3-4sqrt(3)

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2 x^{2} + 9 x = 78 - 3 x

x = - 3 + 43, x = - 3 - 43

Decimal Notation

x = 3.92820 \dots, x = - 9.92820 \dots

Solution steps

2 x^{2} + 9 x = 78 - 3 x

Move

3 x

to the left side

2 x^{2} + 12 x = 78

Move

78

to the left side

2 x^{2} + 12 x - 78 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 78 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 78) = 163

x_{1, 2} = \frac{- 12 \pm 16 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 16 3}{2 \cdot 2}, x_{2} = \frac{- 12 - 16 3}{2 \cdot 2}

x = \frac{- 12 + 16 3}{2 \cdot 2} : - 3 + 43

x = \frac{- 12 - 16 3}{2 \cdot 2} : - 3 - 43

The solutions to the quadratic equation are:

x = - 3 + 43, x = - 3 - 43

\frac{4}{5} (\frac{2}{3} x - \frac{1}{3}) = \frac{2}{5} x - 4

\frac{1}{2} (2) + 1

- x^{2} + x + 1 = 0

5 q + 1 = 6 q - 79

2 x^{2} < x + 6

What are the solutions to the equation 2x^2+9x=78-3x ?
The solutions to the equation 2x^2+9x=78-3x are x=-3+4sqrt(3),x=-3-4sqrt(3)
Find the zeros of 2x^2+9x=78-3x
The zeros of 2x^2+9x=78-3x are x=-3+4sqrt(3),x=-3-4sqrt(3)