(3x-1)(3-x)+4x2+19=0

Solution for (3x-1)(3-x)+4x2+19=0 equation:

(3x-1)(3-x)+4x^2+19=0

We add all the numbers together, and all the variables

4x^2+(3x-1)(-1x+3)+19=0

We multiply parentheses ..

4x^2+(-3x^2+9x+x-3)+19=0

We get rid of parentheses

4x^2-3x^2+9x+x-3+19=0

We add all the numbers together, and all the variables

x^2+10x+16=0

a = 1; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*1}=\frac{-4}{2}
=-2
$