What are the solutions to the equation 4x^2+38x+50=10x-35 ?

The solutions to the equation 4x^2+38x+50=10x-35 are x=-7/2+3i,x=-7/2-3i

Find the zeros of 4x^2+38x+50=10x-35

The zeros of 4x^2+38x+50=10x-35 are x=-7/2+3i,x=-7/2-3i

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4 x^{2} + 38 x + 50 = 10 x - 35

x = - \frac{7}{2} + 3 i, x = - \frac{7}{2} - 3 i

Solution steps

4 x^{2} + 38 x + 50 = 10 x - 35

Move

35

to the left side

4 x^{2} + 38 x + 85 = 10 x

Move

10 x

to the left side

4 x^{2} + 28 x + 85 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 85}{2 \cdot 4}

Simplify

2 8^{2} - 4 \cdot 4 \cdot 85 : 24 i

x_{1, 2} = \frac{- 28 \pm 24 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 24 i}{2 \cdot 4}, x_{2} = \frac{- 28 - 24 i}{2 \cdot 4}

x = \frac{- 28 + 24 i}{2 \cdot 4} : - \frac{7}{2} + 3 i

x = \frac{- 28 - 24 i}{2 \cdot 4} : - \frac{7}{2} - 3 i

The solutions to the quadratic equation are:

x = - \frac{7}{2} + 3 i, x = - \frac{7}{2} - 3 i

x (x + 4) - 6 (x + 4) = 0

2 x^{2} - 2 x = 24

4 x^{2} - 4 x - 17 = 0

x, h = x^{2} + 3 n x + 81 n^{2} + 1

2 x^{2} + 7 = 5 x

What are the solutions to the equation 4x^2+38x+50=10x-35 ?
The solutions to the equation 4x^2+38x+50=10x-35 are x=-7/2+3i,x=-7/2-3i
Find the zeros of 4x^2+38x+50=10x-35
The zeros of 4x^2+38x+50=10x-35 are x=-7/2+3i,x=-7/2-3i