4x(3x+x)+7-5x=8+(-5)(5x-6x)+23

Solution for 4x(3x+x)+7-5x=8+(-5)(5x-6x)+23 equation:

4x(3x+x)+7-5x=8+(-5)(5x-6x)+23

We move all terms to the left:

4x(3x+x)+7-5x-(8+(-5)(5x-6x)+23)=0

We add all the numbers together, and all the variables

4x(+4x)-5x-(8+(-5)(-1x)+23)+7=0

We add all the numbers together, and all the variables

-5x+4x(+4x)-(8+(-5)(-1x)+23)+7=0

We multiply parentheses

16x^2-5x-(8+(-5)(-1x)+23)+7=0

We multiply parentheses ..

16x^2-5x-(8+(+5x)+23)+7=0


We calculate terms in parentheses: -(8+(+5x)+23), so:

8+(+5x)+23

determiningTheFunctionDomain
(+5x)+8+23

We add all the numbers together, and all the variables

(+5x)+31

We get rid of parentheses

5x+31

Back to the equation:

-(5x+31)


We get rid of parentheses

16x^2-5x-5x-31+7=0

We add all the numbers together, and all the variables

16x^2-10x-24=0

a = 16; b = -10; c = -24;
Δ = b2-4ac
Δ = -102-4·16·(-24)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{409}}{2*16}=\frac{10-2\sqrt{409}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{409}}{2*16}=\frac{10+2\sqrt{409}}{32}
$