SOLUTION: Can you please help me with this equation?
3x^2 = 2x+5
So far I got:
3x^2= 2x+5
-3x^2 -3x^2
-3x^2 +2x+5=0
X=-(2) ± (sqr of 2^2 - 4(-3)(5) ) / 2(-3)
X= -2 ± (sqr of 4-
Question 161436: Can you please help me with this equation?
3x^2 = 2x+5
So far I got:
3x^2= 2x+5
-3x^2 -3x^2
-3x^2 +2x+5=0
X=-(2) ± (sqr of 2^2 - 4(-3)(5) ) / 2(-3)
X= -2 ± (sqr of 4- -60) / -6
X= -2 ± (sqr of 64) / -6
X= -2 ± (sqr of 8 )/ -6
X= -2 ± 2 (sqr of 2) /-6
I am unsure what to do after this, if anything at all. Can u show me step by step what to do and explain where I went wrong? I think I'm on the right track but it got tricky with the -6. Thank you so much for even considering my request. Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! 3x^2=2x+5
3x^2-2x-5=0
x=(2+-sqrt[-2^2-4*3*-5])/2*3
x=(2+-sqrt[4+60])/6
x=(2+-sqrt64)/6
x=(2+-8)/6
x=(2+8)/6
x=10/6=5/3 answer.
x=(2-8)/6
x=-6/6
x=-1 answer.