What are the solutions to the equation 49x^2+42x+9=0 ?

The solutions to the equation 49x^2+42x+9=0 are x=-3/7

Find the zeros of 49x^2+42x+9=0

The zeros of 49x^2+42x+9=0 are x=-3/7

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49 x^{2} + 42 x + 9 = 0

x = - \frac{3}{7}

Decimal Notation

x = - 0.42857 \dots

Solution steps

49 x^{2} + 42 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 42 \pm 4 2 ^{2} - 4 \cdot 49 \cdot 9}{2 \cdot 49}

4 2^{2} - 4 \cdot 49 \cdot 9 = 0

x_{1, 2} = \frac{- 42 \pm 0}{2 \cdot 49}

x = \frac{- 42}{2 \cdot 49}

\frac{- 42}{2 \cdot 49} = - \frac{3}{7}

x = - \frac{3}{7}

The solution to the quadratic equation is:

x = - \frac{3}{7}

1 5^{2} + 1 5^{2} = x^{2}

- 2 x^{2} + 4 x - 4 = 0

x^{2} - 13 x = - 5

2 x - (12 x + 6) = 10 x (- 3 x - 7)

(x + 3)^{2} - 5 = 9

What are the solutions to the equation 49x^2+42x+9=0 ?
The solutions to the equation 49x^2+42x+9=0 are x=-3/7
Find the zeros of 49x^2+42x+9=0
The zeros of 49x^2+42x+9=0 are x=-3/7