2(4x2)+7(-3+5x)=(32+44x)

Solution for 2(4x2)+7(-3+5x)=(32+44x) equation:

2(4x^2)+7(-3+5x)=(32+44x)

We move all terms to the left:

2(4x^2)+7(-3+5x)-((32+44x))=0

We add all the numbers together, and all the variables

24x^2+7(5x-3)-((44x+32))=0

We multiply parentheses

24x^2+35x-((44x+32))-21=0


We calculate terms in parentheses: -((44x+32)), so:

(44x+32)

We get rid of parentheses

44x+32

Back to the equation:

-(44x+32)


We get rid of parentheses

24x^2+35x-44x-32-21=0

We add all the numbers together, and all the variables

24x^2-9x-53=0

a = 24; b = -9; c = -53;
Δ = b2-4ac
Δ = -92-4·24·(-53)
Δ = 5169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{5169}}{2*24}=\frac{9-\sqrt{5169}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{5169}}{2*24}=\frac{9+\sqrt{5169}}{48}
$