What are the solutions to the equation 4x^2+85=-36x ?

The solutions to the equation 4x^2+85=-36x are x=-9/2+i,x=-9/2-i

Find the zeros of 4x^2+85=-36x

The zeros of 4x^2+85=-36x are x=-9/2+i,x=-9/2-i

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4 x^{2} + 85 = - 36 x

x = - \frac{9}{2} + i, x = - \frac{9}{2} - i

Solution steps

4 x^{2} + 85 = - 36 x

Move

36 x

to the left side

4 x^{2} + 85 + 36 x = 0

Write in the standard form

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

4 x^{2} + 36 x + 85 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 4 \cdot 85}{2 \cdot 4}

Simplify

3 6^{2} - 4 \cdot 4 \cdot 85 : 8 i

x_{1, 2} = \frac{- 36 \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 36 + 8 i}{2 \cdot 4}, x_{2} = \frac{- 36 - 8 i}{2 \cdot 4}

x = \frac{- 36 + 8 i}{2 \cdot 4} : - \frac{9}{2} + i

x = \frac{- 36 - 8 i}{2 \cdot 4} : - \frac{9}{2} - i

The solutions to the quadratic equation are:

x = - \frac{9}{2} + i, x = - \frac{9}{2} - i

4 k^{2} + 6 k + 1 = 2 k

3 v^{2} + 10 v + 1 = v

y, x^{2} - x y + y^{2} = 0

y, 3 x^{2} + 2 x y + y^{2} = 2

x, c x^{2} + x + y^{2} = 0

What are the solutions to the equation 4x^2+85=-36x ?
The solutions to the equation 4x^2+85=-36x are x=-9/2+i,x=-9/2-i
Find the zeros of 4x^2+85=-36x
The zeros of 4x^2+85=-36x are x=-9/2+i,x=-9/2-i