find the equation 4(H +1)=4?

4(H +1) = 4

To find ''H'' (and hence to solve the equation), we have to get on ''H'' alone in one side of the equation. To do that, first multiply ''4'' outside on the brackets on both the and the ''H'' and the ''1'' inside the brackets. Doing that, we get

4H+4 = 4

Remember, we are still striving to get ''H'' in one side of the equation. To do that, we have to get the rid of the two 4s in the side of the ''H'' so as to have the ''H'' only in that side of the equation. One of the 4s is multiple of ''H'' and the other is added, as we see. First, let us get the red of the added 4. To do that, we need to add -4 (negative 4) (so that the ''4'' will be gone); but, in doing that, we should as well add -4 (negative 4) to the other side of the equation as well because what one side of an equation is done to should be done to the other as well-- otherwise it won't be equal any more, would it? So, adding -4 (negative 4) to both sides of the equation we get

4H+4(-4) = 4+(-4)
4H-4-4=4-4
4H = 0

Now, it should be clear that that H will be zero. But for purposes of understanding how other similar problems should be done, let us go on. Now, we have 4H = 0, and we are still on our way to get the ''H'' on side of the equation, so as to solve it. That means, we need to get the rid of the bloody 4 in front of the ''H''. To do that, we should divide 4H by 4. But remember that you should do as well to the other side of the equation what you do to one side (otherwise they wouldn't be equal any more). That means, in dividing 4H by 4, so as to get the rid of the 4, you should as well divide the ''0'' by 4. That is,

4H = 0
4H/4 =0/4
H = 0/4

Now that we have H on side of the equation, we have solved it. We finally arrived at the golden palace. Hail to us. But wait, what does ''0/4'' mean? In ordinary (real number) arithmetic, ''0/4'' has no meaning. So the answer is 0. One big zero! Our journey was wasted on one big zero in an abandoned castle! Aaah!

H = 0

:)

ok. Say your quadratic formula is 2x^2 - 7x + 6 Set the entire equation to 0. 2x^2 - 7x +6 = 0 Now you ought to look to work out if there is an uncomplicated component. If there is, component it out of the equation, and remedy the equation you have left. because of the fact that there isn't ordinary component in this one, multiply the type final term by applying the type interior the 1st term: (6)(2) = 12 And take the type interior the middle term: -7 Now you ought to use your innovations. you ought to discover 2 numbers that are equivalent to 12 whilst more advantageous and additionally equivalent to -7 whilst further. Multiplication = 12 Addition = -7 consequently, the two numbers are -3 and -4 Now replace -3x and -4x interior the equation, changing -7x 2x^2 -3x - 4x + 6 = 0 group the equation into 2 communities and component each and each group. (2x^2 - 3x) - (4x - 6) = 0 x(2x - 3) - 2(2x - 3) = 0 The binomials you come across interior the brackets ought to consistently be an analogous. in the event that they don't look to be, you have accomplished something incorrect. Now take the term in front of the 1st binomial and upload it to the term in front of the 2d binomial. Multiply this binomial with the different one you discovered. (x-2)(2x - 3) = 0 because of the fact the entire equation is comparable to 0, the two (x - 2) = 0 or (2x - 3) = 0. x - 2 = 0 x = 2 2x -3 = 0 2x = 3 x = 3/2 So hence, whilst y = 0, x = 2 and 3/2. Now, say you have an equation like this : 4x^2 + 5x - 7 = 0 it is surprisingly not uncomplicated to discover the two numbers you ought to replace in for 5x. hence, you utilize the quadratic formula. in case you don't be attentive to it, seem it up. it relatively is not uncomplicated to form it because of the fact that there is no key for sq. root. Use this formula to discover the two x-values and there you have it. here is yet another equation: x^2 - 9 = 0 sq. root each and each term. x and +/- 3 Now multiply x - 3 by applying x + 3 ( x - 3 ) (x + 3) = 0 x - 3 = 0 x = 3 x + 3 = 0 x = -3 x = 3 and x = -3

H = 0

Find the SOLUTION.

4(H + 1) = 4
H + 1 = 4/4
H = 1 - 1
H = 0

4(H +1) = 4
4H + 4= 4
4H=0

=> H=0

develop the product
4(H+1) = 4H+ 4=4
4H=0
H=0

solve for H? H = 0 lol. sorry that seems so basic to me. here are the steps:

4(H + 1) = 4
H + 1 = 1
H = 0

h=0

H =0