*Use the quadratic formula to solve `5x^2+12=-6x` .*

The quadratic formula: If `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)`

(1) First, put into standard form: `5x^2+6x+12=0` (Add 6x to both sides of the equation)

(2) Next identify a,b, and c. a=5,b=6,c=12

(3) Substitute into the formula:

`x=(-6+-sqrt(6^2-4(5)(12)))/(2(5))`

`=(-6+-sqrt(36-240))/10`

`=(-6+-sqrt(-204))/10`

** The square root of a...**

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*Use the quadratic formula to solve `5x^2+12=-6x` .*

The quadratic formula: If `ax^2+bx+c=0` then `x=(-b+-sqrt(b^2-4ac))/(2a)`

(1) First, put into standard form: `5x^2+6x+12=0` (Add 6x to both sides of the equation)

(2) Next identify a,b, and c. a=5,b=6,c=12

(3) Substitute into the formula:

`x=(-6+-sqrt(6^2-4(5)(12)))/(2(5))`

`=(-6+-sqrt(36-240))/10`

`=(-6+-sqrt(-204))/10`

**The square root of a negative number does not exist in the real numbers, so there is no solution in the real numbers.**

** If you are in the complex numbers, the solutions are `-3/5 + (sqrt(51))/5i,-3/5-(sqrt(51))/5i` where `i` is the imaginary number.**