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

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

Find the zeros of 4x^2-2x=3

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

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4 x^{2} - 2 x = 3

x = \frac{1 + 13}{4}, x = \frac{1 - 13}{4}

Decimal Notation

x = 1.15138 \dots, x = - 0.65138 \dots

Solution steps

4 x^{2} - 2 x = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 4 (- 3) = 213

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 2 13}{2 \cdot 4} : \frac{1 + 13}{4}

x = \frac{- ( - 2 ) - 2 13}{2 \cdot 4} : \frac{1 - 13}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{4}, x = \frac{1 - 13}{4}

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

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

5 d^{2} + 15 d + 5 = - 6

2 x^{2} + 4 x - 5

- 8 x^{2} + 4 x = - 5

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