15+4x^2+9x=39+20x

Solution for 15+4x^2+9x=39+20x equation:

15+4x^2+9x=39+20x

We move all terms to the left:

15+4x^2+9x-(39+20x)=0

We add all the numbers together, and all the variables

4x^2+9x-(20x+39)+15=0

We get rid of parentheses

4x^2+9x-20x-39+15=0

We add all the numbers together, and all the variables

4x^2-11x-24=0

a = 4; b = -11; c = -24;
Δ = b2-4ac
Δ = -112-4·4·(-24)
Δ = 505
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{-(-11)-\sqrt{505}}{2*4}=\frac{11-\sqrt{505}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{505}}{2*4}=\frac{11+\sqrt{505}}{8}
$