What are the solutions to the equation 4x^2+12x+9=48 ?

The solutions to the equation 4x^2+12x+9=48 are x=(-3+4sqrt(3))/2 ,x=-(3+4sqrt(3))/2

Find the zeros of 4x^2+12x+9=48

The zeros of 4x^2+12x+9=48 are x=(-3+4sqrt(3))/2 ,x=-(3+4sqrt(3))/2

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4 x^{2} + 12 x + 9 = 48

x = \frac{- 3 + 4 3}{2}, x = - \frac{3 + 4 3}{2}

Decimal Notation

x = 1.96410 \dots, x = - 4.96410 \dots

Solution steps

4 x^{2} + 12 x + 9 = 48

Move

48

to the left side

4 x^{2} + 12 x - 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 39 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 39) = 163

x_{1, 2} = \frac{- 12 \pm 16 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 16 3}{2 \cdot 4}, x_{2} = \frac{- 12 - 16 3}{2 \cdot 4}

x = \frac{- 12 + 16 3}{2 \cdot 4} : \frac{- 3 + 4 3}{2}

x = \frac{- 12 - 16 3}{2 \cdot 4} : - \frac{3 + 4 3}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 4 3}{2}, x = - \frac{3 + 4 3}{2}

168 - 104 x + 12 x^{2} = 0

6 x^{2} - 4 x = 1

x, 5 x^{2} + 4 x y + 2 y^{2} - 24 x - 12 y + 29 = 0

15 a^{2} + 6 a + 3 = 0

m^{2} + 4 m + 13 = 0

What are the solutions to the equation 4x^2+12x+9=48 ?
The solutions to the equation 4x^2+12x+9=48 are x=(-3+4sqrt(3))/2 ,x=-(3+4sqrt(3))/2
Find the zeros of 4x^2+12x+9=48
The zeros of 4x^2+12x+9=48 are x=(-3+4sqrt(3))/2 ,x=-(3+4sqrt(3))/2