What are the roots of the equation?
10x2 - 6x + 4 = -3x

(3 +/- i*sqrt(151) / 20 (no real solutions)

Step-by-step explanation:

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

Add 3x to both sides

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

10x^2 - 6x + 4 + 3x = 0

Add like terms

10x^2 - 3x + 4 = 0

We have to use quadratic formula to find the answers

For this equation, a = 10, b = -3 and c = 4

Note: +/- is plus or minus

Quadratic formula:

x = (-b +/- sqrt(b^2 - 4ac)) / 2a

x = (-(-3) +/- sqrt(-3^2 - 4(10)(4))) / 2(10)

x = (3 +/- sqrt(9 - 160)) / 20

x = (3 +/- sqrt(-151) / 20

x = (3 +/- i*sqrt(151) / 20

if there's a "-" sign inside a square root, it will give you an imaginary number.

Example:

-sqrt(9) = -3

sqrt(-9) = 3i (first you will ignore the "- " sign, then square the number. After finding the square root, take that '-' sign and turn it to an "i")

So for sqrt(-151), 151 is not a perfect square, so you will get a decimal number. Leave it as it is:

sqrt(151). We can't forget about the "-". Turn the minus into a sign.

It will be i*sqrt(151)

Imaginary numbers can't be simplified, so this is the answer.

(3 +/- i*sqrt(151) / 20

the solution is 12x-6y

step-by-step explanation: