(9x-4)+(8x2)=(13x+30)+(19x-11)

Solution for (9x-4)+(8x2)=(13x+30)+(19x-11) equation:

(9x-4)+(8x^2)=(13x+30)+(19x-11)

We move all terms to the left:

(9x-4)+(8x^2)-((13x+30)+(19x-11))=0

determiningTheFunctionDomain
8x^2+(9x-4)-((13x+30)+(19x-11))=0

We get rid of parentheses

8x^2+9x-((13x+30)+(19x-11))-4=0


We calculate terms in parentheses: -((13x+30)+(19x-11)), so:

(13x+30)+(19x-11)

We get rid of parentheses

13x+19x+30-11

We add all the numbers together, and all the variables

32x+19

Back to the equation:

-(32x+19)


We get rid of parentheses

8x^2+9x-32x-19-4=0

We add all the numbers together, and all the variables

8x^2-23x-23=0

a = 8; b = -23; c = -23;
Δ = b2-4ac
Δ = -232-4·8·(-23)
Δ = 1265
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{-(-23)-\sqrt{1265}}{2*8}=\frac{23-\sqrt{1265}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1265}}{2*8}=\frac{23+\sqrt{1265}}{16}
$