How do you solve #y=2x^2+5x-1# using the quadratic formula?

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Answer 1 · 2016-03-02T15:56:43

I will start you on the way so that you can complete the calculation

Given:#" "y=2x^2+5x-1#........................(1)

By solve I assume you mean solve for:#" "2x^2+5x-1=0#

Standard form equation: #" "y=ax^2bx+c#

Where:#" "x=(-b+-sqrt(b^2-4ac))/(2a)# ............(2)

and
#a=2#
#b=5#
#c=(-1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitution in equation (2) gives

#x=(-5+-sqrt(5^2-4(2)(-1)))/(2(2))#

#(-5+-sqrt(25+8))/(4)#

#color(blue)("I will let you finish this bit. There should be 2 answers")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

You did not ask for vertex coordinates or x-intercept or y-intercept so not included!