What are the solutions to the equation 4x^2+12x=-3 ?

The solutions to the equation 4x^2+12x=-3 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2

Find the zeros of 4x^2+12x=-3

The zeros of 4x^2+12x=-3 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2

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4 x^{2} + 12 x = - 3

x = \frac{- 3 + 6}{2}, x = - \frac{3 + 6}{2}

Decimal

x = - 0.27525 \dots, x = - 2.72474 \dots

Solution steps

4 x^{2} + 12 x = - 3

Move

3

to the left side

4 x^{2} + 12 x + 3 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 4 \cdot 3 = 46

x_{1, 2} = \frac{- 12 \pm 4 6}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 4 6}{2 \cdot 4}, x_{2} = \frac{- 12 - 4 6}{2 \cdot 4}

x = \frac{- 12 + 4 6}{2 \cdot 4} : \frac{- 3 + 6}{2}

x = \frac{- 12 - 4 6}{2 \cdot 4} : - \frac{3 + 6}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 6}{2}, x = - \frac{3 + 6}{2}

2 x^{2} = 8

x^{2} + 4 x + 14 = 46

(20 - x) (10 - x) = 0

3 x^{2} - 7 = 20

v^{2} - 6 v + 5 = 0

What are the solutions to the equation 4x^2+12x=-3 ?
The solutions to the equation 4x^2+12x=-3 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2
Find the zeros of 4x^2+12x=-3
The zeros of 4x^2+12x=-3 are x=(-3+sqrt(6))/2 ,x=-(3+sqrt(6))/2