Question

how would solve quadratic equations like x^2+2x-1=2 and
5r^2=80

Answer 1

5r^2=80

r^2=80/5

Answer 2

To solve quadratic equation u should use the formula :-
\[\LARGE{\color{green}{-b \pm \sqrt{b^2-4ac}\over2a}}\]

Answer 3

so would r^2=80/5 be the answer or the first step

Answer 4

\[\LARGE{5r^2=80}\]\[\LARGE{r^2=\frac{80}{5}}\]\[\LARGE{r=\frac{\sqrt{80}}{\sqrt{5}}}\]

Answer 5

First:
x^2+2x-1=2
x^2+2x-1-2=2-2
x^2+2x-3=0
Factor the expression:
(x+3)(x-1) = 0
Put each factor equal to 0 and solve them. That is to solve:
x+3 = 0 or x-1 = 0

Second one:
Divide both sides by 5 first
5r^2 = 80
r^2 = 80/5 = 16
So, r^2 = 16
Take square root
r= \(\pm \sqrt{16}\) = ...

Answer 6

If you want to use the quadratic formula mentioned in the above, make sure you've got the equation in the form \(ax^2 + bx + c = 0\)

Answer 7

yup right :)

Answer 8

@mopishmold2575 r^2 = 80/5 is only the first step.

Answer 9

ok but how would you put it in the quadratic form since there 4 numbers
youd have a number for the a variable the b variable and the c variable but what would you do with the fourth variable

Answer 10

wait i got it thanks