3x^2+16x-64=1-x^2

Solution for 3x^2+16x-64=1-x^2 equation:

3x^2+16x-64=1-x^2

We move all terms to the left:

3x^2+16x-64-(1-x^2)=0

We get rid of parentheses

3x^2+x^2+16x-1-64=0

We add all the numbers together, and all the variables

4x^2+16x-65=0

a = 4; b = 16; c = -65;
Δ = b2-4ac
Δ = 162-4·4·(-65)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-36}{2*4}=\frac{-52}{8}
=-6+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+36}{2*4}=\frac{20}{8}
=2+1/2
$