3x^2+5x-20=16+x-x^2

Solution for 3x^2+5x-20=16+x-x^2 equation:

3x^2+5x-20=16+x-x^2

We move all terms to the left:

3x^2+5x-20-(16+x-x^2)=0

We get rid of parentheses

3x^2+x^2-x+5x-16-20=0

We add all the numbers together, and all the variables

4x^2+4x-36=0

a = 4; b = 4; c = -36;
Δ = b2-4ac
Δ = 42-4·4·(-36)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{37}}{2*4}=\frac{-4-4\sqrt{37}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{37}}{2*4}=\frac{-4+4\sqrt{37}}{8}
$