What are the solutions to the equation 10x^2-6x+4=-3x ?

The solutions to the equation 10x^2-6x+4=-3x are x= 3/20+i(sqrt(151))/(20),x= 3/20-i(sqrt(151))/(20)

Find the zeros of 10x^2-6x+4=-3x

The zeros of 10x^2-6x+4=-3x are x= 3/20+i(sqrt(151))/(20),x= 3/20-i(sqrt(151))/(20)

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10 x^{2} - 6 x + 4 = - 3 x

x = \frac{3}{20} + i \frac{151}{20}, x = \frac{3}{20} - i \frac{151}{20}

Solution steps

10 x^{2} - 6 x + 4 = - 3 x

Move

3 x

to the left side

10 x^{2} - 3 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 10 \cdot 4}{2 \cdot 10}

Simplify

(- 3)^{2} - 4 \cdot 10 \cdot 4 : 151 i

x_{1, 2} = \frac{- ( - 3 ) \pm 151 i}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 151 i}{2 \cdot 10}, x_{2} = \frac{- ( - 3 ) - 151 i}{2 \cdot 10}

x = \frac{- ( - 3 ) + 151 i}{2 \cdot 10} : \frac{3}{20} + i \frac{151}{20}

x = \frac{- ( - 3 ) - 151 i}{2 \cdot 10} : \frac{3}{20} - i \frac{151}{20}

The solutions to the quadratic equation are:

x = \frac{3}{20} + i \frac{151}{20}, x = \frac{3}{20} - i \frac{151}{20}

x^{2} + 30 x + 225 = 0

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

x^{2} + 7 x + 30 = 0

p^{2} - 37 p = 0

0 = (x + 1) (x - 3)

What are the solutions to the equation 10x^2-6x+4=-3x ?
The solutions to the equation 10x^2-6x+4=-3x are x= 3/20+i(sqrt(151))/(20),x= 3/20-i(sqrt(151))/(20)
Find the zeros of 10x^2-6x+4=-3x
The zeros of 10x^2-6x+4=-3x are x= 3/20+i(sqrt(151))/(20),x= 3/20-i(sqrt(151))/(20)