Factor the following quadratic equation;
d="TexFormula1" title="12 {x}^{2} + 22x - 14" alt="12 {x}^{2} + 22x - 14" align="absmiddle" class="latex-formula">
1 answer:
Answer:
2(3x + 7)(2x - 1)
Step-by-step explanation:
You can see it a little easier if you take out a common factor of 2
2(6x^2 + 11x - 7)
The 6 leaves you with a lot of factors, the 7 does not. It only has 2 factors.
Let 6 factor into 2 and 3 and the 7 into 7 and 1
2(3x - 1 )(2x + 7)
Now remove the brackets.
2(6x^2 + 21x - 2x - 7) This obviously does not work but we'll combine like terms anyway.
2(6x^2 + 19x - 7)
So we'll try it again
2(3x + 7)(2x - 1)
2(6x^2 + 14x - 3x - 7) Looks like we have it.
2(6x^2 + 11x - 7)
So the right factors are
2(3x + 7)(2x - 1)
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Step-by-step explanation:




![$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$](https://tex.z-dn.net/?f=%24%24The%20domain%20of%20%24s%24%20is%20%24%5B0%2C19%5D%24.So%20the%20endpoints%20are%200%20and%2019%24%24%5Cbegin%7Baligned%7D%26A_%7BT%2BS%7D%280%29%3D%5Cfrac%7B0%5E%7B2%7D%7D%7B16%7D%2B%5Cfrac%7B%5Csqrt%7B3%7D%2819-0%29%5E%7B2%7D%7D%7B36%7D%20%5Capprox%2017.36%20%5C%5C%26A_%7BT%2BS%7D%288.26%29%3D%5Cfrac%7B8.26%5E%7B2%7D%7D%7B16%7D%2B%5Cfrac%7B%5Csqrt%7B3%7D%2819-8.26%29%5E%7B2%7D%7D%7B36%7D%20%5Capprox%209.81%20%5C%5C%26A_%7BT%2BS%7D%2819%29%3D%5Cfrac%7B19%5E%7B2%7D%7D%7B16%7D%2B%5Cfrac%7B%5Csqrt%7B3%7D%2819-19%29%5E%7B2%7D%7D%7B36%7D%3D22.56%5Cend%7Baligned%7D%24%24)

Answer:
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where you want to talk??
i mean which app?
I don't see a picture, sorry...
The two lines are parallel.