What are the solutions to the equation 5x^2+29x-42=0 ?

The solutions to the equation 5x^2+29x-42=0 are x= 6/5 ,x=-7

Find the zeros of 5x^2+29x-42=0

The zeros of 5x^2+29x-42=0 are x= 6/5 ,x=-7

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5 x^{2} + 29 x - 42 = 0

x = \frac{6}{5}, x = - 7

Decimal Notation

x = 1.2, x = - 7

Solution steps

5 x^{2} + 29 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 29 \pm 2 9 ^{2} - 4 \cdot 5 ( - 42 )}{2 \cdot 5}

2 9^{2} - 4 \cdot 5 (- 42) = 41

x_{1, 2} = \frac{- 29 \pm 41}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 29 + 41}{2 \cdot 5}, x_{2} = \frac{- 29 - 41}{2 \cdot 5}

x = \frac{- 29 + 41}{2 \cdot 5} : \frac{6}{5}

x = \frac{- 29 - 41}{2 \cdot 5} : - 7

The solutions to the quadratic equation are:

x = \frac{6}{5}, x = - 7

\frac{2 5 - 2}{5 + 2}

\frac{4}{3} \cdot 9

(3 k^{4})^{4}

3 x^{2} - 9 = 7 x

3 x - 2 + 2 x + 5 = 3 (x + 5)

What are the solutions to the equation 5x^2+29x-42=0 ?
The solutions to the equation 5x^2+29x-42=0 are x= 6/5 ,x=-7
Find the zeros of 5x^2+29x-42=0
The zeros of 5x^2+29x-42=0 are x= 6/5 ,x=-7