How do you solve the equation 2x^2+5x-3=02x2+5x3=0 by using the quadratic formula?

1 Answer
Dec 26, 2016

The two solutions are x = 0.5x=0.5 and x = -3x=3

Explanation:

Since this question is given in standard form, meaning that it follows the form: ax^(2) + bx + c = 0ax2+bx+c=0, we can use the quadratic formula to solve for x:

http://www.1728.org/quadratc.htmhttp://www.1728.org/quadratc.htm
I think it's worthwhile to mention that aa is the number that has the x^2x2 term associated with it. Thus, it would be 2x^(2)2x2 for this question.bb is the number that has the xx variable associated with it and it would be 5x5x, and cc is a number by itself and in this case it is -3.

Now, we just plug our values into the equation like this:

x = (- (5) +- sqrt((5)^(2) - 4(2)(-3)))/(2(2))x=(5)±(5)24(2)(3)2(2)

x = (-5 +-sqrt(25+24))/4x=5±25+244

x = (-5 +- sqrt(49))/4x=5±494

For these type of problems, you will obtain two solutions because of the +-± part. So what you want to do is add -5 and sqrt(49)49 together and divide that by 4:

x = (-5+sqrt(49))/4x=5+494
x = 2/4= 0.5x=24=0.5

Now, we subtract sqrt(49)49 from -5 and divide by 4:

x = (-5-sqrt(49))/4x=5494
x = -12/4 = -3x=124=3

Therefore, the two possible solutions are:
x = 0.5x=0.5 and x = -3x=3