Solving equation with quadratic formula. Question will be given below

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Solving equation with quadratic formula. Question will be given below

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\large 2x^2 - 5x = 52\]
I know that the quadratic formula is: \[\large x = \frac{ -b \pm \sqrt{b^2 - 2ac} }{ 2a }\]
a quadratic equation of the form: ax^2 + bx + c = 0 has roots defined by the quadratic formula as: \[x=\frac{1}{2a}(-b\pm\sqrt{b^2-4ac})\] which is just a shortcut of completing the square

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\large x = \frac{ -(-5) \pm \sqrt{(-5)^2 - 2(2)(-52)} }{ 2(2) }\] if i solve it like this will it be fine?
might be a typo, but its 4ac, not 2ac
otherwise your values for abc are good
ok
\[x = \frac{ 5 \pm \sqrt {25 + 208} }{ 4 }\]
if solve it with this will it be correct? @amistre64

Not the answer you are looking for?

Search for more explanations.

Ask your own question