How do you find the zeros, real and imaginary, of y= -7x^2-5x+3y=−7x2−5x+3 using the quadratic formula?

1 Answer
The Legend
Feb 21, 2018

The zeroes (or roots, as they are also called) of y=-7x^2-5x+3y=−7x2−5x+3 are x=(5+-sqrt(109))/-14x=5±√109−14

Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a. aa,bb, and cc come from ax^2+bx+cax2+bx+c, the standard form of a quadratic equation.

The equation is already in standard form (thank goodness). a=-7a=−7, b=-5b=−5, and c=3c=3.
Plugging these into the quadratic formula:
x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a
x=(-(-5)+-sqrt((-5)^2-4(-7)(3)))/(2(-7))x=−(−5)±√(−5)2−4(−7)(3)2(−7)
x=(5+-sqrt(25-(-28)(3)))/-14x=5±√25−(−28)(3)−14
x=(5+-sqrt(25-(-84)))/-14x=5±√25−(−84)−14
x=(5+-sqrt(25+84))/-14x=5±√25+84−14
x=(5+-sqrt(109))/-14x=5±√109−14
109 is a prime number, so sqrt(109)√109 is in simplest form. Therefore our answer is x=(5+-sqrt(109))/-14x=5±√109−14

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