What are the solutions to the equation -3x^2-7x+5=0 ?

The solutions to the equation -3x^2-7x+5=0 are x=-(7+sqrt(109))/6 ,x=(sqrt(109)-7)/6

Find the zeros of -3x^2-7x+5=0

The zeros of -3x^2-7x+5=0 are x=-(7+sqrt(109))/6 ,x=(sqrt(109)-7)/6

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- 3 x^{2} - 7 x + 5 = 0

x = - \frac{7 + 109}{6}, x = \frac{109 - 7}{6}

Decimal Notation

x = - 2.90671 \dots, x = 0.57338 \dots

Solution steps

- 3 x^{2} - 7 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 3 ) \cdot 5}{2 ( - 3 )}

(- 7)^{2} - 4 (- 3) \cdot 5 = 109

x_{1, 2} = \frac{- ( - 7 ) \pm 109}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 109}{2 ( - 3 )}, x_{2} = \frac{- ( - 7 ) - 109}{2 ( - 3 )}

x = \frac{- ( - 7 ) + 109}{2 ( - 3 )} : - \frac{7 + 109}{6}

x = \frac{- ( - 7 ) - 109}{2 ( - 3 )} : \frac{109 - 7}{6}

The solutions to the quadratic equation are:

x = - \frac{7 + 109}{6}, x = \frac{109 - 7}{6}

6 y - 6 y^{2} = 0

7 x - 3 x^{2} = - 10

1009 = - 2 t^{2} + 108 t + 1

5.6 \cdot 1 0^{- 10} = \frac{x ^{2}}{0.5}

j (t) = - \frac{1}{3} t^{2} + 6 t - 10

What are the solutions to the equation -3x^2-7x+5=0 ?
The solutions to the equation -3x^2-7x+5=0 are x=-(7+sqrt(109))/6 ,x=(sqrt(109)-7)/6
Find the zeros of -3x^2-7x+5=0
The zeros of -3x^2-7x+5=0 are x=-(7+sqrt(109))/6 ,x=(sqrt(109)-7)/6