(3x-1)(x+2)=x2+16

Solution for (3x-1)(x+2)=x2+16 equation:

(3x-1)(x+2)=x2+16

We move all terms to the left:

(3x-1)(x+2)-(x2+16)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

-x^2+(+3x^2+6x-1x-2)-16=0

We add all the numbers together, and all the variables

-1x^2+(+3x^2+6x-1x-2)-16=0

We get rid of parentheses

-1x^2+3x^2+6x-1x-2-16=0

We add all the numbers together, and all the variables

2x^2+5x-18=0

a = 2; b = 5; c = -18;
Δ = b2-4ac
Δ = 52-4·2·(-18)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*2}=\frac{-18}{4}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*2}=\frac{8}{4}
=2
$