4(2x-3)=6x+2
Please can someone give me the working out and answer?

The roots are

x=±√3,1±√6i

Step-by-step explanation:

(x2+a)(x2bx+c)=0

Multiply the two quadratics.

x4+bx3+(a+c)x2+abx+ac=0

Comparing terms gives.

b=−2,a+c=4,ab=6,ac=−21

We have

b=−2, ab=6

means that

a=−3.

now:

a+c=4 means c=7.

Verify with the last term

ac=−3⋅7=−21.

So the equation factorises to.

(x2−3)(x2−2x+7)=0

Factorise each term to give the roots.

x2−3=0, is x2=3, roots x=±√3x2−2x+7=0

complete the square

(x−1)2=−6

gives complex roots

x=±√6i

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step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.