(x^2-16)+(3x)=90

Solution for (x^2-16)+(3x)=90 equation:

(x^2-16)+(3x)=90

We move all terms to the left:

(x^2-16)+(3x)-(90)=0

We add all the numbers together, and all the variables

3x+(x^2-16)-90=0

We get rid of parentheses

x^2+3x-16-90=0

We add all the numbers together, and all the variables

x^2+3x-106=0

a = 1; b = 3; c = -106;
Δ = b2-4ac
Δ = 32-4·1·(-106)
Δ = 433
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{-(3)-\sqrt{433}}{2*1}=\frac{-3-\sqrt{433}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{433}}{2*1}=\frac{-3+\sqrt{433}}{2}
$