What are the solutions to the equation 5x^2-3x-6=0 ?

The solutions to the equation 5x^2-3x-6=0 are x=(3+sqrt(129))/(10),x=(3-sqrt(129))/(10)

Find the zeros of 5x^2-3x-6=0

The zeros of 5x^2-3x-6=0 are x=(3+sqrt(129))/(10),x=(3-sqrt(129))/(10)

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5 x^{2} - 3 x - 6 = 0

x = \frac{3 + 129}{10}, x = \frac{3 - 129}{10}

Decimal Notation

x = 1.43578 \dots, x = - 0.83578 \dots

Solution steps

5 x^{2} - 3 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 6 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 6) = 129

x_{1, 2} = \frac{- ( - 3 ) \pm 129}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 129}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 129}{2 \cdot 5}

x = \frac{- ( - 3 ) + 129}{2 \cdot 5} : \frac{3 + 129}{10}

x = \frac{- ( - 3 ) - 129}{2 \cdot 5} : \frac{3 - 129}{10}

The solutions to the quadratic equation are:

x = \frac{3 + 129}{10}, x = \frac{3 - 129}{10}

\frac{x - 6}{4} + 3 = \frac{x}{10}

x^{3} + 6 x^{2} < 13 x + 42

∣ 5 - 6 x ∣ = 7

y^{2} - 18 = 0

13 x^{4} y^{2} + 39 x^{5} y

What are the solutions to the equation 5x^2-3x-6=0 ?
The solutions to the equation 5x^2-3x-6=0 are x=(3+sqrt(129))/(10),x=(3-sqrt(129))/(10)
Find the zeros of 5x^2-3x-6=0
The zeros of 5x^2-3x-6=0 are x=(3+sqrt(129))/(10),x=(3-sqrt(129))/(10)