I need to solve by completing the square:?

Question

x^2 = 5x + 2

I came up with 5 +- sqrt29/2

Is that correct?


I couldnt find an answer for this one:

Solve by completing the square as well:

4x^2 + 2x - 3 =0

Answer 1

ill write down the formula for finding the root of a quadratic equation : ax^2 + bx + c = 0
solution: x = {-b +- sqrt(b^2 - 4ac) } / 2a

equation x^2=5x + 2 can be written as :

x^2 - 5x - 2 = 0
here a = 1 ; b = -5 ; c = -2 ;
simply apply the formula to find the value of x .

x ={ -(-5)+-sqrt(25 - 4*-2) } / 2
x ={ 5 +- sqrt(24+8) } / 2
x = { 5 +- sqrt(33) } / 2

The second equation :

4x^2 + 2x - 3 = 0
here a = 4 ; b = 2 ; c = -3

x = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4
x = {-2 +- sqrt(4+48) } / 8
x = { -2 +- sqrt(52) } /8

simplify the term inside the square root , 52 = 13*4
so 4 gets square rooted and

x = { -2 +- 2 sqrt(13) } /8

take 2 common from the numerator and denominator and cancel them.

x = 2{ -1 +- sqrt(13) } / 4*2
x = { -1 +- sqrt(13) } / 4

thats your answer.

have a nice day :)

Answer 2

x^2=5x+2
=>x^2-5x=2
=>(x)^2-2.x.(5/2)+(5/2)^2=2+(5/2)^2 [adding (5/5)^2 in both sides
=>(x-5/2)^2=2+25/4=33/4
=>x-5/2=[(sqrt)+-33]/2 [square-rooting both sides
=>x=(5/2)+-[(sqrt)33]/2=[ 5+-sqrt33]/2 ans

4x^2+2x-3=0
=>(2x)^2+2.2x.(1/2)+(1/2)^2-(1/2)^2-3=0
=>(2x+1/2)^2=3+(1/4)=13/4
=>(2x+1/2)=(+-sqrt13)/2
=>2x=(+-sqrt13)/2-(1/2)=[+-(sqrt13)-1]/ 2
=>x=[+-(sqrt13)-1]/4 ans

Answer 3

for the second one:
4x^2 + 2x - 3 =0
x(2x+2)-3=0
x(2x+2)=3
x=3/(2x+2)

i dunno if this is the right answer