Solving quadratic equations

Divide by 2 first and take constant to the RHS:  $x^2+3x=\frac{5}{2}$

Now add square of half the coefficient of x to both sides:  $x^2+3x+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}$

Rewrite as:  $(x+\frac{3}{2})^2=\frac{19}{4}$

You should be able to take it from here.

Remeber the 'Quadratic Formula' ???   Use it.....

Nope

Nobody said anything about a formula unless you mean "Take half he coefficient of x and square it and add it to both sides"

Do you do that before or after dividing by 2?

$\text{let s use } (a+b)^2=a^2+2ab+b^2\text{ and }a^2-b^2=(a-b)(a+b):\\x^2 + 3x - 2.5=0\\x^2 + 3x - 2.5+4.75=0+4.75\\x^2 + 3x + 2.25=4.75\text{ (1.5*1.5=2.25 1.5*2=3x)}\\(x+1.5)^2=4.75\\(x+1.5)^2-4.75=0\\(x+1.5-\sqrt{4.75})*(x+1.5+\sqrt{4.75})=0\\x_1=0.679449\\x_2=-3.67945$

FOLLOW THE PROCEDURE OUTLINED IN THIS SOLUTION:

Solve for x:

2 x^2+6 x-5 = 0

Divide both sides by 2:

x^2+3 x-5/2 = 0

Add 5/2 to both sides:

x^2+3 x = 5/2

Add 9/4 to both sides:

x^2+3 x+9/4 = 19/4

Write the left hand side as a square:

(x+3/2)^2 = 19/4

Take the square root of both sides:

x+3/2 = sqrt(19)/2 or x+3/2 = -sqrt(19)/2

Subtract 3/2 from both sides:

x = sqrt(19)/2-3/2 or x+3/2 = -sqrt(19)/2

Subtract 3/2 from both sides:

Answer: |x = sqrt(19)/2-3/2                or x = -3/2-sqrt(19)/2