All categories

3 years ago

Will mark brainliest!! solve & simplify the following equation: 2x²-7x=4

Mathematics

1 answer:

Goryan [66]3 years ago

4 0

Answer:

x= −1 /2 or x=4

Step-by-step explanation:

This is ALGEBRA, equations, factoring and zero property. ok.. let's solve this step-by-step

2x^2−7x=4

Step 1: Subtract 4 from both sides.

2x^2−7x−4 =4 −4

2x^2−7x−4=0

Step 2: Factor left side of equation.

(2x+1)(x−4)=0

Step 3: Set factors equal to 0.

2x+1=0 or x−4=0

x= −1 /2 or x=4

(BTW, there are supposed to be two answers, it's factoring)

Sorry if the format is not clear, but I hope this helped (really feel you, algebra can be brain-striking)!! Please leave a comment, rating and/or mark as the brainliest-whichever you think I deserve! Thank You :)

You might be interested in

How many times does 1/3 go into 1/18?

Orlov [11]

6 is the answer 18 divided by 3 equals to 6 so that’s the answer

4 0

3 years ago

Read 2 more answers

NEED HELP MY GRADE DEPENDS ON THIS!!!

ss7ja [257]

28x - 2 = 26x + 6
2x = 8
x = 4

4 0

3 years ago

(X-5)1/2+5=2 what is a possible solution?

Dmitry [639]

Answer:

X=-1

Step-by-step explanation:

(X-5)1/2+5=2

(X-5)1/2=2-5

(X-5)1/2=-3

(X-5)=-3 *2

X-5=-6

X=-1

6 0

3 years ago

A rectangular vegetable garden will have a width that is feet less than the length, and an area of square feet. If x represents

ivanzaharov [21]

8(8-2)=48
8(6)=48
therefore x=8

5 0

2 years ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$