What are the solutions to the equation 4x^2+6x=1 ?

The solutions to the equation 4x^2+6x=1 are x=(-3+sqrt(13))/4 ,x=-(3+sqrt(13))/4

Find the zeros of 4x^2+6x=1

The zeros of 4x^2+6x=1 are x=(-3+sqrt(13))/4 ,x=-(3+sqrt(13))/4

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4 x^{2} + 6 x = 1

x = \frac{- 3 + 13}{4}, x = - \frac{3 + 13}{4}

Decimal Notation

x = 0.15138 \dots, x = - 1.65138 \dots

Solution steps

4 x^{2} + 6 x = 1

Move

1

to the left side

4 x^{2} + 6 x - 1 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 4 (- 1) = 213

x_{1, 2} = \frac{- 6 \pm 2 13}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 6 + 2 13}{2 \cdot 4}, x_{2} = \frac{- 6 - 2 13}{2 \cdot 4}

x = \frac{- 6 + 2 13}{2 \cdot 4} : \frac{- 3 + 13}{4}

x = \frac{- 6 - 2 13}{2 \cdot 4} : - \frac{3 + 13}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 13}{4}, x = - \frac{3 + 13}{4}

- 92

340

x^{2} - 4 x + 10 = 0

3 (2 x - 3) + 2 (2 x - 2) + 10 = 5 x + 17

y, \frac{l y l}{2} = 28

What are the solutions to the equation 4x^2+6x=1 ?
The solutions to the equation 4x^2+6x=1 are x=(-3+sqrt(13))/4 ,x=-(3+sqrt(13))/4
Find the zeros of 4x^2+6x=1
The zeros of 4x^2+6x=1 are x=(-3+sqrt(13))/4 ,x=-(3+sqrt(13))/4