What are the solutions to the equation 4x^2-8x=3 ?

The solutions to the equation 4x^2-8x=3 are x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2

Find the zeros of 4x^2-8x=3

The zeros of 4x^2-8x=3 are x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2

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4 x^{2} - 8 x = 3

x = \frac{2 + 7}{2}, x = \frac{2 - 7}{2}

Decimal Notation

x = 2.32287 \dots, x = - 0.32287 \dots

Solution steps

4 x^{2} - 8 x = 3

Move

3

to the left side

4 x^{2} - 8 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 (- 3) = 47

x_{1, 2} = \frac{- ( - 8 ) \pm 4 7}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 7}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 4 7}{2 \cdot 4}

x = \frac{- ( - 8 ) + 4 7}{2 \cdot 4} : \frac{2 + 7}{2}

x = \frac{- ( - 8 ) - 4 7}{2 \cdot 4} : \frac{2 - 7}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 7}{2}, x = \frac{2 - 7}{2}

- \frac{4}{4}

2 x^{2} - 17 x y + 15 y^{2}

5 - 5 x + 7 x = 8 - 12 x

- 7 > 3 + 2 k

\frac{1}{2} + 2 = 4

What are the solutions to the equation 4x^2-8x=3 ?
The solutions to the equation 4x^2-8x=3 are x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2
Find the zeros of 4x^2-8x=3
The zeros of 4x^2-8x=3 are x=(2+sqrt(7))/2 ,x=(2-sqrt(7))/2