# solve x^2=3x+5must show work

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The equation to be solved is x^2 = 3x + 5

x^2 = 3x + 5

=> x^2 - 3x - 5 = 0

`x1 = (3 + sqrt (9 + 20))/2`

=> `3/2 + sqrt 29`

`x2 = 3/2 - sqrt 29`

**The roots of the equation x^2 = 3x + 5 are `1.5 + sqrt 29` and **`1.5 - sqrt 29`

First take all the numbers to one side so u get a equation like this :-

X^2-3X-5=0

Then go on solving.But how??????

Remeber these two equations so when you get questions like this its easy to solve, the equation is:-

(-b-sqrt(b^2-4ac))/2a and -b+sqrt(b^2-4ac))/2a

then go on subtituting the values as :- a=1 (when no value is in front of x itz always 1)

b=-3

c=-5

Then go on suibstituing the values for both the equationns. u will end up getting two answers as :- X= 4.2 (1dp)

X= -1.2(1dp)

these two will be your final answer.

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