3x+4=64-16x+x^2

Solution for 3x+4=64-16x+x^2 equation:

3x+4=64-16x+x^2

We move all terms to the left:

3x+4-(64-16x+x^2)=0

We get rid of parentheses

-x^2+16x+3x-64+4=0

We add all the numbers together, and all the variables

-1x^2+19x-60=0

a = -1; b = 19; c = -60;
Δ = b2-4ac
Δ = 192-4·(-1)·(-60)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*-1}=\frac{-30}{-2}
=+15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*-1}=\frac{-8}{-2}
=+4
$