What are the solutions to the equation 4x^2-6x-8=0 ?

The solutions to the equation 4x^2-6x-8=0 are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4

Find the zeros of 4x^2-6x-8=0

The zeros of 4x^2-6x-8=0 are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4

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4 x^{2} - 6 x - 8 = 0

x = \frac{3 + 41}{4}, x = \frac{3 - 41}{4}

Decimal Notation

x = 2.35078 \dots, x = - 0.85078 \dots

Solution steps

4 x^{2} - 6 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ( - 8 )}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 (- 8) = 241

x_{1, 2} = \frac{- ( - 6 ) \pm 2 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 41}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 2 41}{2 \cdot 4}

x = \frac{- ( - 6 ) + 2 41}{2 \cdot 4} : \frac{3 + 41}{4}

x = \frac{- ( - 6 ) - 2 41}{2 \cdot 4} : \frac{3 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{4}, x = \frac{3 - 41}{4}

\frac{4}{6} - \frac{1}{2}

3 (2 x - 3)^{2} = 4 x (2 x - 9) + 43

\frac{- \frac{1}{12}}{- \frac{1}{12} + 1} - 1

18 x^{2} + 24 x = 0

x^{3} - 4 x^{2} + 9 x - 10

What are the solutions to the equation 4x^2-6x-8=0 ?
The solutions to the equation 4x^2-6x-8=0 are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4
Find the zeros of 4x^2-6x-8=0
The zeros of 4x^2-6x-8=0 are x=(3+sqrt(41))/4 ,x=(3-sqrt(41))/4