What are the solutions to the equation 4x^2+12x=-13 ?

The solutions to the equation 4x^2+12x=-13 are x=-3/2+i,x=-3/2-i

Find the zeros of 4x^2+12x=-13

The zeros of 4x^2+12x=-13 are x=-3/2+i,x=-3/2-i

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4 x^{2} + 12 x = - 13

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

Solution steps

4 x^{2} + 12 x = - 13

Move

13

to the left side

4 x^{2} + 12 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

Simplify

1 2^{2} - 4 \cdot 4 \cdot 13 : 8 i

x_{1, 2} = \frac{- 12 \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 12 + 8 i}{2 \cdot 4}, x_{2} = \frac{- 12 - 8 i}{2 \cdot 4}

x = \frac{- 12 + 8 i}{2 \cdot 4} : - \frac{3}{2} + i

x = \frac{- 12 - 8 i}{2 \cdot 4} : - \frac{3}{2} - i

The solutions to the quadratic equation are:

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

\frac{1}{2} (\frac{3}{2})

x^{3} - 3 x^{2} + 5 x - 15

- 3 x - 2 \geq 7

\frac{8 - 4 x}{3} - 2 (5 x + 8) = \frac{2 ( 4 x + 6 )}{9} + 2 (10 x + 1)

3^{- 2} = x

What are the solutions to the equation 4x^2+12x=-13 ?
The solutions to the equation 4x^2+12x=-13 are x=-3/2+i,x=-3/2-i
Find the zeros of 4x^2+12x=-13
The zeros of 4x^2+12x=-13 are x=-3/2+i,x=-3/2-i