16-x^2=3x+12

Solution for 16-x^2=3x+12 equation:

16-x^2=3x+12

We move all terms to the left:

16-x^2-(3x+12)=0

We add all the numbers together, and all the variables

-1x^2-(3x+12)+16=0

We get rid of parentheses

-1x^2-3x-12+16=0

We add all the numbers together, and all the variables

-1x^2-3x+4=0

a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2}
=-4
$