What are the solutions to the equation 2x^2-8x-3=0 ?

The solutions to the equation 2x^2-8x-3=0 are x=(4+sqrt(22))/2 ,x=(4-sqrt(22))/2

Find the zeros of 2x^2-8x-3=0

The zeros of 2x^2-8x-3=0 are x=(4+sqrt(22))/2 ,x=(4-sqrt(22))/2

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

x = \frac{4 + 22}{2}, x = \frac{4 - 22}{2}

Decimal Notation

x = 4.34520 \dots, x = - 0.34520 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 (- 3) = 222

x_{1, 2} = \frac{- ( - 8 ) \pm 2 22}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 22}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 2 22}{2 \cdot 2}

x = \frac{- ( - 8 ) + 2 22}{2 \cdot 2} : \frac{4 + 22}{2}

x = \frac{- ( - 8 ) - 2 22}{2 \cdot 2} : \frac{4 - 22}{2}

The solutions to the quadratic equation are:

x = \frac{4 + 22}{2}, x = \frac{4 - 22}{2}

3 \leq 2 x - 5 < 7

25000

(x - 2)^{2} + (0 + 5)^{2} = 9

\frac{1}{4} x^{4} + x^{2} + 1

1 6^{2} + 3 0^{2} = c^{2}

What are the solutions to the equation 2x^2-8x-3=0 ?
The solutions to the equation 2x^2-8x-3=0 are x=(4+sqrt(22))/2 ,x=(4-sqrt(22))/2
Find the zeros of 2x^2-8x-3=0
The zeros of 2x^2-8x-3=0 are x=(4+sqrt(22))/2 ,x=(4-sqrt(22))/2