Question

solve the equation x2+3x-5=0

Answer 1

you have to use the quadratic formula

Answer 2

Observe that this is not a perfect square
=> meaning cross method wouldn't work

So if there are 2 ways to solve it now:
M1- Completing the Square
M2- Graphics calc [APPS -> Plysmlt -> poly root finder]

M1: *note that coefficient of x^2 must always be one
\[( x ^{2} + 3x ) -5 = 0 \]
\[(x + 3/2 )^{2} - (3/2)^{2} - 5 = 0 \]
\[(x + 3/2)^{2} -(29/4) \]

Answer 3

Bascially completing the sq of ax2 + bx + c = 0
a(x2 + bx) + c = 0
a (x + b/2)^2 - a(b/2)^2 + c = 0

*note that if a is not 1, the value that you minus out (b/2)^2 must be a(b/2)^2!