2x(squared)+x-3=0 5x(squared)+3x=2 x(squared)-5x=0 5x(squared)= 4x these are all separate equations, can somebody solve them? !

[Note: ^2 is squared. i just find it quicker to type out]

2x^2 + x - 3 = 0
First get rid of -3 by adding 3 to both sides.
2x^2 + x = +3
Then get rid of x by subtracting x from both sides
2x^2 = 3 - x
Now square root (√) both sides to get rid of the square (^2). This leaves 2x by itself.
2x = √(3 - x)
Finally, divide both sides by 2 so x can be by itself.
x = √(3 - x)/2

5x^2 + 3x =2
Similar to question the 1st question. Get rid of -3x because its smaller than 5x^2
5x^2 = 2 + 3x
Now get rid of the square by sq rooting both sides
5x = √(2 -3x)
This time you divide both sides by 5
x = √(2 - 3x)/5

x^2 - 5x = 0
+5x to both sides
x^2 = 5x
sq root both sides to remove ^2
x = √(5x)
I think this part is optional since you've already got x by itself, but you could divde both sidee by 5 so its not 5x
x/5 = √x

I'll say b) log(- that if not let me know what i can do