What are the solutions to the equation 9x^2-3x-5=0 ?

The solutions to the equation 9x^2-3x-5=0 are x=(1+sqrt(21))/6 ,x=(1-sqrt(21))/6

Find the zeros of 9x^2-3x-5=0

The zeros of 9x^2-3x-5=0 are x=(1+sqrt(21))/6 ,x=(1-sqrt(21))/6

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

x = \frac{1 + 21}{6}, x = \frac{1 - 21}{6}

Decimal Notation

x = 0.93042 \dots, x = - 0.59709 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 ( - 5 )}{2 \cdot 9}

(- 3)^{2} - 4 \cdot 9 (- 5) = 321

x_{1, 2} = \frac{- ( - 3 ) \pm 3 21}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 21}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 3 21}{2 \cdot 9}

x = \frac{- ( - 3 ) + 3 21}{2 \cdot 9} : \frac{1 + 21}{6}

x = \frac{- ( - 3 ) - 3 21}{2 \cdot 9} : \frac{1 - 21}{6}

The solutions to the quadratic equation are:

x = \frac{1 + 21}{6}, x = \frac{1 - 21}{6}

∣ 4 x - 7 ∣ = 21

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

\frac{2 ( \frac{4}{3} )}{1 - ( \frac{4}{3} ) ^{2}}

\frac{1}{3} + 4

x^{2} + 484 = 0

What are the solutions to the equation 9x^2-3x-5=0 ?
The solutions to the equation 9x^2-3x-5=0 are x=(1+sqrt(21))/6 ,x=(1-sqrt(21))/6
Find the zeros of 9x^2-3x-5=0
The zeros of 9x^2-3x-5=0 are x=(1+sqrt(21))/6 ,x=(1-sqrt(21))/6