What are the solutions to the equation 3x^1+4x^2=3 ?

The solutions to the equation 3x^1+4x^2=3 are x=(-3+sqrt(57))/8 ,x=(-3-sqrt(57))/8

Find the zeros of 3x^1+4x^2=3

The zeros of 3x^1+4x^2=3 are x=(-3+sqrt(57))/8 ,x=(-3-sqrt(57))/8

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3 x^{1} + 4 x^{2} = 3

x = \frac{- 3 + 57}{8}, x = \frac{- 3 - 57}{8}

Decimal Notation

x = 0.56872 \dots, x = - 1.31872 \dots

Solution steps

3 x^{1} + 4 x^{2} = 3

Refine

3 x + 4 x^{2} = 3

Move

3

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 4 (- 3) = 57

x_{1, 2} = \frac{- 3 \pm 57}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 3 + 57}{2 \cdot 4}, x_{2} = \frac{- 3 - 57}{2 \cdot 4}

x = \frac{- 3 + 57}{2 \cdot 4} : \frac{- 3 + 57}{8}

x = \frac{- 3 - 57}{2 \cdot 4} : \frac{- 3 - 57}{8}

The solutions to the quadratic equation are:

x = \frac{- 3 + 57}{8}, x = \frac{- 3 - 57}{8}

(\frac{1}{10})^{5}

x^{2} = 5

- 6 f + 24 \leq 60 and f - 14 \leq - 22

5 (6 y - 9) = - 15

(x + 5)^{2} - 6 = 0

What are the solutions to the equation 3x^1+4x^2=3 ?
The solutions to the equation 3x^1+4x^2=3 are x=(-3+sqrt(57))/8 ,x=(-3-sqrt(57))/8
Find the zeros of 3x^1+4x^2=3
The zeros of 3x^1+4x^2=3 are x=(-3+sqrt(57))/8 ,x=(-3-sqrt(57))/8