# Solve 3 Quadratic equations.. 1. 10x^2 + 4x + 2 = 02. 4x^2 + 12x + 10 = 03. x^2 + 3x - 9 = 0

1) `10x^2+4x+2 = 0`

First try to simplify this by dividing by 2. This would get,

`5x^2+2x+1 = 0`

To solve this problem you can use the general formula.

For a quadratic equation of `ax^2+bx+c = 0` , the solutions are,

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

Using this formula for this,

`x = (-2+-sqrt(2^2-4(5)(1)))/(2(5))`

`x = (-2+-sqrt(4-20))/10`

`x = (-2+-sqrt(-16))/10`

Now this would give complex roots,

`x = (-2+-4i)/10`

Therefore two solutions are `x = (-1+2i)/5` and `x = (-1-2i)/5`

2) `4x^2+12x+10 = 0`

This also can be simplified by dividing by 2.

`2x^2+6x+5 = 0`

The same procedure can be used here too,

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

`x = (-6+-sqrt(6^2-4(2)(5)))/(2(2))`

`x = (-6+-sqrt(36-40))/4`

`x = (-6+-sqrt(-4))/4`

This equations also has complex roots.

`x = (-6+-2i)/4`

Therefore the two solutions are, `x = (-3+i)/2` and `x = (-3-i)/2`

3)`x^2+3x-9 = 0`

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

`x = (-3+-sqrt(3^2-4(1)(-9)))/(2(1))`

`x = (-3+-sqrt(9+36))/2`

`x = (-3+-sqrt(45))/2`

`x = (-3+-3sqrt(5))/2`

Therefore the two solutions are, `x = (-3+3sqrt(5))/2` and `x = (-3-3sqrt(5))/2`

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