# How do you solve #7d ^ { 2} + 10d + 168= 0#?

##### 2 Answers

#### Explanation:

We can find the complex roots by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with

#0 = 7(7d^2+10d+168)#

#color(white)(0) = 49d^2+70d+1176#

#color(white)(0) = 49d^2+70d+25+1151#

#color(white)(0) = (7d)^2+2(7d)(5)+(5)^2+1151#

#color(white)(0) = (7d+5)^2+(sqrt(1151))^2#

#color(white)(0) = (7d+5)^2-(sqrt(1151)i)^2#

#color(white)(0) = ((7d+5)-sqrt(1151)i)((7d+5)+sqrt(1151)i)#

#color(white)(0) = (7d+5-sqrt(1151)i)(7d+5+sqrt(1151)i)#

So:

#7d = -5+-sqrt(1151)i#

and:

#d = -5/7+-sqrt(1151)/7 i#

#### Explanation:

we use this formula

in this case our

putting them into equation gives us

by solving we get