What are the solutions to the equation quadraticformula 4x^2+9x+3=0 ?

The solutions to the equation quadraticformula 4x^2+9x+3=0 are x=(-9+sqrt(33))/8 ,x=(-9-sqrt(33))/8

Find the zeros of quadraticformula 4x^2+9x+3=0

The zeros of quadraticformula 4x^2+9x+3=0 are x=(-9+sqrt(33))/8 ,x=(-9-sqrt(33))/8

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4 x^{2} + 9 x + 3 = 0

x = \frac{- 9 + 33}{8}, x = \frac{- 9 - 33}{8}

Decimal Notation

x = - 0.40692 \dots, x = - 1.84307 \dots

Solution steps

4 x^{2} + 9 x + 3 = 0

Solve with the quadratic formula

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

9^{2} - 4 \cdot 4 \cdot 3 = 33

x_{1, 2} = \frac{- 9 \pm 33}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 9 + 33}{2 \cdot 4}, x_{2} = \frac{- 9 - 33}{2 \cdot 4}

x = \frac{- 9 + 33}{2 \cdot 4} : \frac{- 9 + 33}{8}

x = \frac{- 9 - 33}{2 \cdot 4} : \frac{- 9 - 33}{8}

The solutions to the quadratic equation are:

x = \frac{- 9 + 33}{8}, x = \frac{- 9 - 33}{8}

X, X^{2} + Y^{2} = 0

- 6 x (x - 1) = 0

x^{2} - 39 = 0

r, (x - 2)^{2} + y^{2} = r^{2}

lo g_{10} (2) (x^{2} - 4) - lo g_{10} (2) (2 x + 4) = 4

What are the solutions to the equation quadraticformula 4x^2+9x+3=0 ?
The solutions to the equation quadraticformula 4x^2+9x+3=0 are x=(-9+sqrt(33))/8 ,x=(-9-sqrt(33))/8
Find the zeros of quadraticformula 4x^2+9x+3=0
The zeros of quadraticformula 4x^2+9x+3=0 are x=(-9+sqrt(33))/8 ,x=(-9-sqrt(33))/8