What are the solutions to the equation 3x^2+15x-42=0 ?

The solutions to the equation 3x^2+15x-42=0 are x=2,x=-7

Find the zeros of 3x^2+15x-42=0

The zeros of 3x^2+15x-42=0 are x=2,x=-7

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3 x^{2} + 15 x - 42 = 0

x = 2, x = - 7

Solution steps

3 x^{2} + 15 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 3 ( - 42 )}{2 \cdot 3}

1 5^{2} - 4 \cdot 3 (- 42) = 27

x_{1, 2} = \frac{- 15 \pm 27}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 15 + 27}{2 \cdot 3}, x_{2} = \frac{- 15 - 27}{2 \cdot 3}

x = \frac{- 15 + 27}{2 \cdot 3} : 2

x = \frac{- 15 - 27}{2 \cdot 3} : - 7

The solutions to the quadratic equation are:

x = 2, x = - 7

4 y + 2 = 10

\frac{( 3 m + 1 ) ( m - 1 )}{( m + 1 ) ^{2}} > 0

2 (10 a - 13) + 5 a = 619

20 + 5

1 \cdot 1 0^{- 5}

What are the solutions to the equation 3x^2+15x-42=0 ?
The solutions to the equation 3x^2+15x-42=0 are x=2,x=-7
Find the zeros of 3x^2+15x-42=0
The zeros of 3x^2+15x-42=0 are x=2,x=-7