# Find the roots via factoring

I'm trying to solve a quadratic by factoring, the equation given is $$3{x}^{2} + 5x = 6$$

The first thing I did to try and solve it is by putting the $$6$$ on to the other side of the equation making it $$3{x}^{2} + 5x - 6$$ Though, trying to solve it now by looking at some common factor and, it looks like it's not even possible to factor properly. Any help with this?

• I think the roots are real, but not rational. Aug 4, 2023 at 16:31
• @ChrisLewis The roots are real. Aug 4, 2023 at 16:31
• Double checking and, no, this is how it was shown exactly. I'm not really sure how I'm expected to factor this down without using another method like completing the square or the quadratic formula. But thank you for double checking for me, I will now go check with my teacher regarding this. Aug 4, 2023 at 16:31
• Clearly it is solvable by the quadratic equation with two real (and irrational) roots, so you're not going to be able to 'factor' it into nice clean integer roots or rational roots. Perhaps solve by "completing the square" if the exercise doesn't allow the use of the quadratic equation. Aug 4, 2023 at 16:33
• Multiply by $4\cdot 3$. You get $(2\cdot 3\cdot x)^2+2\cdot (2\cdot 3\cdot x)\cdot 5=4\cdot 3\cdot 6$. Now add $5^2$. You get $(2\cdot 3\cdot x+5)^2=4\cdot 3\cdot 6+5^2=(\sqrt{4\cdot 3\cdot 6+5^2})^2$. Now you have a difference of squares.
– NDB
Aug 4, 2023 at 16:34

For any quadratic equation $$cx^2+a_1x+b_1=0$$ with $$c \neq 0$$ ($$c = 0$$ leads to a linear equation) we can divide both sides by $$c$$ to get $$x^2+ax+b=0$$

In general for any equation $$x^2+ax+b=0$$: $$0=x^2+ax+b=x^2+ax+(\frac{a}{2})^2-(\frac{a}{2})^2 +b= (x+\frac{a}{2})^2-(\frac{a}{2})^2 +b$$ so we obtain

$$(x+\frac{a}{2})^2-((\frac{a}{2})^2-b)=0$$

Now factor as a difference of squares if possible

• Why the downvote?
– Sgg8
Aug 6, 2023 at 11:17
• I upvoted, but it may be helpful if you were to add to your answer a bit more. In particular, convert the OP's equation $3x^2+5x=6$ into an equation of the form $x^2+ax+b=0$.
– Mike
Aug 15, 2023 at 17:28
• I edited. Fine now?
– Sgg8
Aug 17, 2023 at 9:01
• It looks like basically the same answer to me. You didn't take my suggestion in my above comment? [I don't think it matter too much anyway as the people following this question have probably moved on.]
– Mike
Aug 17, 2023 at 17:29
• I explained how to get rid of the leading term coefficient
– Sgg8
Aug 17, 2023 at 18:10