so it asks you to use the quadratic formula which is when you have ax^2+bx+c=0 then you put it into the quadratic formula which is x=((-b+-√(b^2-4ac))/(2a))
so 8x^2+3x=-7
add 7 to both sides
8x^2+3x+7=0
put into the quadratic formula
8x^2+3x+7=0
ax^2+bx+c=0
x=((-b+-√(b^2-4ac))/(2a))
x=((-3+-√(3^2-4(8)(7)))/(2(8)))
solve inside of square root first
3^2-4(8)(7)=pemdas 9-4(8)(7)=9-224=-215
so -3+-√-215
we know that square root of negative is not real and is immaginary so i must be involved
√-1=i so
-215=i√215
so the answer is A
CONTINUATION
i√215=i(15)=15i
-3+-15i means -3+ or minus 15i
then solve for botom part
2(8)=16
(-3+-15i)/16
so the answers are (-3+15i)/16 and (-3-15i)/16
