Question 240628
This is a job for the quadratic formula. For
{{{ax^2 + bx + c = 0}}}
the solutions can be found with:
{{{x = (-b +- sqrt(b^2 - 4ac))/(2a)}}}<br>
In your equation, {{{x^2-5x+35}}}
a = 1
b = -5
c = 35
So the solutions are:
{{{x = (-(-5) +- sqrt((-5)^2 - 4(1)(35)))/(2(1))}}}
Simplifying inside the square root:
{{{x = (-(-5) +- sqrt(25 - 4(1)(35)))/(2(1))}}}
{{{x = (-(-5) +- sqrt(25 - 140))/(2(1))}}}
{{{x = (-(-5) +- sqrt(-115))/(2(1))}}}
At this point, since there is a negative number in the square root, we know that the solutions will be complex (not irrational). For complex numbers we use "i" which stand for {{{sqrt(-1)}}}:
{{{x = (-(-5) +- sqrt((-1)(115)))/(2(1))}}}
{{{x = (-(-5) +- sqrt(-1)sqrt(115))/(2(1))}}}
{{{x = (-(-5) +- i*sqrt(115))/(2(1))}}}
Simplifying the rest we get:
{{{x = (5 +- i*sqrt(115))/2}}}
Separating this into two terms and writing this in a+bi form we get:
{{{x = 5/2 +- (sqrt(115)/2)i}}}