What are the solutions to the equation 4x^2+29=-20x ?

The solutions to the equation 4x^2+29=-20x are x=-5/2+i,x=-5/2-i

Find the zeros of 4x^2+29=-20x

The zeros of 4x^2+29=-20x are x=-5/2+i,x=-5/2-i

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4 x^{2} + 29 = - 20 x

x = - \frac{5}{2} + i, x = - \frac{5}{2} - i

Solution steps

4 x^{2} + 29 = - 20 x

Move

20 x

to the left side

4 x^{2} + 29 + 20 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} + 20 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

Simplify

2 0^{2} - 4 \cdot 4 \cdot 29 : 8 i

x_{1, 2} = \frac{- 20 \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 8 i}{2 \cdot 4}, x_{2} = \frac{- 20 - 8 i}{2 \cdot 4}

x = \frac{- 20 + 8 i}{2 \cdot 4} : - \frac{5}{2} + i

x = \frac{- 20 - 8 i}{2 \cdot 4} : - \frac{5}{2} - i

The solutions to the quadratic equation are:

x = - \frac{5}{2} + i, x = - \frac{5}{2} - i

(2 + 3 i)^{2} + (2 - 3 i)^{2}

2 x - 3 (x + 5) = 2 x - (x + 3)

\frac{6 ^{\frac{1}{2}}}{6 ^{\frac{1}{3}}}

(\frac{( 30.03 + 0.21 )}{30.42}) - 1

x^{2} + 2 x = - 4

What are the solutions to the equation 4x^2+29=-20x ?
The solutions to the equation 4x^2+29=-20x are x=-5/2+i,x=-5/2-i
Find the zeros of 4x^2+29=-20x
The zeros of 4x^2+29=-20x are x=-5/2+i,x=-5/2-i