(13x^2-19)=(9x^+6)

Solution for (13x^2-19)=(9x^+6) equation:

(13x^2-19)=(9x^+6)

We move all terms to the left:

(13x^2-19)-((9x^+6))=0

We get rid of parentheses

13x^2-((9x^+6))-19=0


We calculate terms in parentheses: -((9x^+6)), so:

(9x^+6)

We get rid of parentheses

9x^+6

We add all the numbers together, and all the variables

9x+6

Back to the equation:

-(9x+6)


We get rid of parentheses

13x^2-9x-6-19=0

We add all the numbers together, and all the variables

13x^2-9x-25=0

a = 13; b = -9; c = -25;
Δ = b2-4ac
Δ = -92-4·13·(-25)
Δ = 1381
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{1381}}{2*13}=\frac{9-\sqrt{1381}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{1381}}{2*13}=\frac{9+\sqrt{1381}}{26}
$