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Find the unknown.$$\frac{3}{8}(2 r+7)^{2}+\frac{15}{2}=\frac{3}{4}$$

$$\frac{-7 \pm 3 \sqrt{2 i}}{2}$$

Algebra

Chapter 0

Reviewing the Basics

Section 2

Solving Equations of the Form $a x^{2}-b=0$

Equations and Inequalities

Oregon State University

Baylor University

Lectures

02:10

Find the unknown.$$\fr…

02:52

00:52

Find the unknown.$$r^{…

01:54

Find the unknown.$$(4 …

02:00

Find the unknown.$$(2 …

01:24

Find the value of each exp…

00:36

Multiply.$$(8 r+2) 7$$…

01:48

01:01

Divide.$$\frac{r^{3}+8…

01:26

Perform the indicated oper…

with this problem, I would start by getting rid of these fractions because we need to get our by itself. In order to get our by itself, we need to get r squared by itself. Um, equals three force. So I'm starting with the least common multiple of those denominators, which is eight. So what I can do is multiply the left and the right side by eight. And as I distribute in here, Well, first of all, those eights will cancel. And here eight divided by two, is four times 15 60. And on this side eight, divided by four is two times three is six. So now my goal is to get this quantity squared by itself. I can do that by subtracting 60 over, and I'm getting six to buy. My negative 60 is negative. 54. But then I also need to get rid of this multiplied by three by dividing by three. So, on the left side, I still have that to r plus seven squared. And on the right side, I get negative. 18. So now I can solve the square part by square rooting. Um And so the square root in the square cancel. I'm left with two R plus seven and on the right side. I know the I for the negative number, and we can rewrite 18 Is Route nine Times group to I should be in there as well because the square to nine is three. So three I moved to, um and so we're almost ready to solve. For are all I have to do is subtract that seven over from nineties three, I wrote to, but I start to get rid of this, too. So to solve a multiplication problem, you divide that to over and we've finished. That's your final answer.

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