How do you solve #2x^2 +5x=3#?

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Answer 1 · 2016-04-05T09:52:21

#x=-3" or "1/2#

Given:#" "2x^2+5x=3#
Write as:#" "2x^2+5x-3=0#

The whole number factors of 3 are 1 and 3.
The whole number factors of 2 are 1 and 2.

Attempt 1

#(2x+3)(x+1) = 2x^2+2x+3x+3color(red)(" Fail")#

Attempt 2

#(2x+1)(x-3) = 2x^2-6x+x-3color(red)(" Fail")#

Looks as though there is at least 1 non whole number solution to this so use the formula

#y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)#

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

#x=(-5+-sqrt(25+24))/4#

#x=(-5+-sqrt(49))/4#

#x=(-5+-7)/4#

#x=-3" or "1/2#