(10x-4)(3x+8)=-122x-33

Solution for (10x-4)(3x+8)=-122x-33 equation:

(10x-4)(3x+8)=-122x-33

We move all terms to the left:

(10x-4)(3x+8)-(-122x-33)=0

We get rid of parentheses

(10x-4)(3x+8)+122x+33=0

We multiply parentheses ..

(+30x^2+80x-12x-32)+122x+33=0

We get rid of parentheses

30x^2+80x-12x+122x-32+33=0

We add all the numbers together, and all the variables

30x^2+190x+1=0

a = 30; b = 190; c = +1;
Δ = b2-4ac
Δ = 1902-4·30·1
Δ = 35980
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{35980}=\sqrt{4*8995}=\sqrt{4}*\sqrt{8995}=2\sqrt{8995}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(190)-2\sqrt{8995}}{2*30}=\frac{-190-2\sqrt{8995}}{60}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(190)+2\sqrt{8995}}{2*30}=\frac{-190+2\sqrt{8995}}{60}
$