What are the solutions to the equation 3x+(2x)^2-10=20 ?

The solutions to the equation 3x+(2x)^2-10=20 are x=(-3+sqrt(489))/8 ,x=(-3-sqrt(489))/8

Find the zeros of 3x+(2x)^2-10=20

The zeros of 3x+(2x)^2-10=20 are x=(-3+sqrt(489))/8 ,x=(-3-sqrt(489))/8

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3 x + (2 x)^{2} - 10 = 20

x = \frac{- 3 + 489}{8}, x = \frac{- 3 - 489}{8}

Decimal Notation

x = 2.38916 \dots, x = - 3.13916 \dots

Solution steps

3 x + (2 x)^{2} - 10 = 20

Expand

3 x + (2 x)^{2} - 10 : 3 x + 4 x^{2} - 10

3 x + 4 x^{2} - 10 = 20

Move

20

to the left side

4 x^{2} + 3 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 4 ( - 30 )}{2 \cdot 4}

3^{2} - 4 \cdot 4 (- 30) = 489

x_{1, 2} = \frac{- 3 \pm 489}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 489}{2 \cdot 4}, x_{2} = \frac{- 3 - 489}{2 \cdot 4}

x = \frac{- 3 + 489}{2 \cdot 4} : \frac{- 3 + 489}{8}

x = \frac{- 3 - 489}{2 \cdot 4} : \frac{- 3 - 489}{8}

The solutions to the quadratic equation are:

x = \frac{- 3 + 489}{8}, x = \frac{- 3 - 489}{8}

x, x^{2} \cdot y + x \cdot y^{2} = 6

1 x^{2} + 6 x = 9

y, 18 y^{2} = (x - 4) \cdot (x - 10)^{2}

7 x^{2} - 12 x + 18 = 0

a^{2} \cdot (x - b) - b^{2} \cdot (x - a) = x^{2} (a - b)

What are the solutions to the equation 3x+(2x)^2-10=20 ?
The solutions to the equation 3x+(2x)^2-10=20 are x=(-3+sqrt(489))/8 ,x=(-3-sqrt(489))/8
Find the zeros of 3x+(2x)^2-10=20
The zeros of 3x+(2x)^2-10=20 are x=(-3+sqrt(489))/8 ,x=(-3-sqrt(489))/8