# Solve the equation (u+7)^2 = 2u^2 +8u +42?

(u+7)^2 = 2u^2 +8u +42

### 7 Answers

- Anonymous1 decade agoBest Answer
(u+7)^2 = 2u^2 +8u +42

u^2+14u+49=2u^2+8u+42

0=u^2-6u-7

0=(u-7)(u+1)

u-7=0 or u+1=0

u=7 or -1

- 1 decade ago
expand the eqn:

(u+7)^2 = 2u^2 +8u +42

u^2 + 14u + 49 = 2u^2 +8u +42

u^2 - 6u - 7 = 0

factorise:

(u-7)(u+1) = 0

therefore u = 7 or u = -1

- robertonereoLv 41 decade ago
(u+7)^2 = 2u^2 +8u +42

u^2 + 14u + 49 = 2u^2 + 8u + 42

2u^2 - u^2 + 8u - 14u + 42 - 49 = 0

u^2 - 6u - 7 = 0

u = [6+- Sq Rt (36 + 28)]/2

u = [6 +- (Sq Rt 64)]/2

u = (6 + 8)/2 = 7

u = (6 - 8)/2 = -1

- fonnerLv 43 years ago
we've a linear equation here. The question is calling you: what style circumstances 7 = 40 two? the plain answer is 6, good? So, x = 6...selection B to resolve it the algebra way, divide the two aspects of the equation via 7. 7x = 40 two x = 40 two divided via 7 x = 6

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- 1 decade ago
u=Sqrt(2u^2+8u+42)-7...u^2=2u^2+8u+42+49....0=u^2+8u+91....use quadratic equation...i think this is right i could be wrong

- 1 decade ago
There are tons of similar questions pasted all over the board. Try to understand them.