Question

solve using quad formula

2x^2+12x=-7

Answer 1

We have the equation:
\[2x^2+12x=-7\]
How the quadratic formula works is:

"Give me a equation in the form of \(ax^2+bc+c=0\) and I will tell you that the roots, (or the solutions) of the equation are exactly:
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)"

So we need to re-arrange the equation \(2x^2+12x=-7\) into that form. It's really not that bad as adding 7 to both sides will result in: \(2x^2+12x+7=0\).

So now that we have it in that form, lets identify our variables. We have the variables x, a, b, and c. We have a formula which will give us x if we give it a, b, and c. So lets identify those three variables. We can tell just by looking at it that:
\[\eqalign{
&a=2\\
&b=12\\
&c=7\\
}\]
So put those into the formula:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
And tell me what you get! :)

Answer 2

i did and i got

x= -12√200
/
4

i couldnt type it as good as you did but u get what im saying anyways thats were im stuck

Answer 3

Well lets see:
\[\eqalign{
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\
&=\frac{-12\pm\sqrt{144-4(2)(7)}}{4} \\
&=\frac{-12\pm\sqrt{144-64}}{4} \\
&=\frac{-12\pm\sqrt{80}}{4} \\
&=\frac{-12\pm\sqrt{16\times5}}{4} \\
&=\frac{-12\pm\sqrt{16}\sqrt{5}}{4} \\
&=\frac{-12\pm4\sqrt{5}}{4} \\
&=-3\pm\sqrt{5} \\
}\]
So that means that our answers are either:
\[x=-3+\sqrt{5}\approx0.76\phantom{......}or\phantom{......}x=-3-\sqrt{5}\approx-5.23\]

Answer 4

Err correction...the first one is \(-0.76\), not \(0.76\)

Answer 5

Thanks a lot! i cant believe how simple of a mistake i was making and i couldnt notice it! that what happens when you study and are sleepy lol

Answer 6

Haha I know eh? Anyways, no problem!