4x(2x+1)=27+3(2x-5)

Solution for 4x(2x+1)=27+3(2x-5) equation:

4x(2x+1)=27+3(2x-5)

We move all terms to the left:

4x(2x+1)-(27+3(2x-5))=0

We multiply parentheses

8x^2+4x-(27+3(2x-5))=0


We calculate terms in parentheses: -(27+3(2x-5)), so:

27+3(2x-5)

determiningTheFunctionDomain
3(2x-5)+27

We multiply parentheses

6x-15+27

We add all the numbers together, and all the variables

6x+12

Back to the equation:

-(6x+12)


We get rid of parentheses

8x^2+4x-6x-12=0

We add all the numbers together, and all the variables

8x^2-2x-12=0

a = 8; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·8·(-12)
Δ = 388
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{97}}{2*8}=\frac{2-2\sqrt{97}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{97}}{2*8}=\frac{2+2\sqrt{97}}{16}
$