What are the solutions to the equation 3x^2+9x=10 ?

The solutions to the equation 3x^2+9x=10 are x=(-9+sqrt(201))/6 ,x=(-9-sqrt(201))/6

Find the zeros of 3x^2+9x=10

The zeros of 3x^2+9x=10 are x=(-9+sqrt(201))/6 ,x=(-9-sqrt(201))/6

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3 x^{2} + 9 x = 10

x = \frac{- 9 + 201}{6}, x = \frac{- 9 - 201}{6}

Decimal Notation

x = 0.86290 \dots, x = - 3.86290 \dots

Solution steps

3 x^{2} + 9 x = 10

Move

10

to the left side

3 x^{2} + 9 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 ( - 10 )}{2 \cdot 3}

9^{2} - 4 \cdot 3 (- 10) = 201

x_{1, 2} = \frac{- 9 \pm 201}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 201}{2 \cdot 3}, x_{2} = \frac{- 9 - 201}{2 \cdot 3}

x = \frac{- 9 + 201}{2 \cdot 3} : \frac{- 9 + 201}{6}

x = \frac{- 9 - 201}{2 \cdot 3} : \frac{- 9 - 201}{6}

The solutions to the quadratic equation are:

x = \frac{- 9 + 201}{6}, x = \frac{- 9 - 201}{6}

3 x^{2} + 2 + 4 x^{2} = - 8 x

a (x) = x \frac{1}{3} (x + 4)

0 = 7 + k^{2} + 8 k

(r^{2} - 4) = 0

x^{2} - 3 x = x + 5

What are the solutions to the equation 3x^2+9x=10 ?
The solutions to the equation 3x^2+9x=10 are x=(-9+sqrt(201))/6 ,x=(-9-sqrt(201))/6
Find the zeros of 3x^2+9x=10
The zeros of 3x^2+9x=10 are x=(-9+sqrt(201))/6 ,x=(-9-sqrt(201))/6