3(4+5x)+8x-2(3x)=1+5x(2-4x^)+20x^

Solution for 3(4+5x)+8x-2(3x)=1+5x(2-4x^)+20x^ equation:

3(4+5x)+8x-2(3x)=1+5x(2-4x^)+20x^

We move all terms to the left:

3(4+5x)+8x-2(3x)-(1+5x(2-4x^)+20x^)=0

We add all the numbers together, and all the variables

3(5x+4)+8x-23x-(1+5x(2-4x^)+20x^)=0

We add all the numbers together, and all the variables

-15x+3(5x+4)-(1+5x(2-4x^)+20x^)=0

We multiply parentheses

-15x+15x-(1+5x(2-4x^)+20x^)+12=0


We calculate terms in parentheses: -(1+5x(2-4x^)+20x^), so:

1+5x(2-4x^)+20x^

determiningTheFunctionDomain
5x(2-4x^)+20x^+1

We add all the numbers together, and all the variables

20x+5x(2-4x^)+1

We multiply parentheses

-20x^2+20x+10x+1

We add all the numbers together, and all the variables

-20x^2+30x+1

Back to the equation:

-(-20x^2+30x+1)


We add all the numbers together, and all the variables

-(-20x^2+30x+1)+12=0

We get rid of parentheses

20x^2-30x-1+12=0

We add all the numbers together, and all the variables

20x^2-30x+11=0

a = 20; b = -30; c = +11;
Δ = b2-4ac
Δ = -302-4·20·11
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{5}}{2*20}=\frac{30-2\sqrt{5}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{5}}{2*20}=\frac{30+2\sqrt{5}}{40}
$