What are the solutions to the equation 3x^2+20x+36=4 ?

The solutions to the equation 3x^2+20x+36=4 are x=-8/3 ,x=-4

Find the zeros of 3x^2+20x+36=4

The zeros of 3x^2+20x+36=4 are x=-8/3 ,x=-4

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3 x^{2} + 20 x + 36 = 4

x = - \frac{8}{3}, x = - 4

Decimal Notation

x = - 2.66666 \dots, x = - 4

Solution steps

3 x^{2} + 20 x + 36 = 4

Move

4

to the left side

3 x^{2} + 20 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 3 \cdot 32}{2 \cdot 3}

2 0^{2} - 4 \cdot 3 \cdot 32 = 4

x_{1, 2} = \frac{- 20 \pm 4}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 20 + 4}{2 \cdot 3}, x_{2} = \frac{- 20 - 4}{2 \cdot 3}

x = \frac{- 20 + 4}{2 \cdot 3} : - \frac{8}{3}

x = \frac{- 20 - 4}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

x = - \frac{8}{3}, x = - 4

10 x^{2} - 16 x + 6 = 0

r, v = \frac{4}{3} π r^{2}

9 x^{2} - 54 x

x^{2} + 19 x + 46 = 0

3 x^{2} + 113 x - 7 = 0

What are the solutions to the equation 3x^2+20x+36=4 ?
The solutions to the equation 3x^2+20x+36=4 are x=-8/3 ,x=-4
Find the zeros of 3x^2+20x+36=4
The zeros of 3x^2+20x+36=4 are x=-8/3 ,x=-4