(2x-3)(x+4)-8x=10

Solution for (2x-3)(x+4)-8x=10 equation:

(2x-3)(x+4)-8x=10

We move all terms to the left:

(2x-3)(x+4)-8x-(10)=0

We add all the numbers together, and all the variables

-8x+(2x-3)(x+4)-10=0

We multiply parentheses ..

(+2x^2+8x-3x-12)-8x-10=0

We get rid of parentheses

2x^2+8x-3x-8x-12-10=0

We add all the numbers together, and all the variables

2x^2-3x-22=0

a = 2; b = -3; c = -22;
Δ = b2-4ac
Δ = -32-4·2·(-22)
Δ = 185
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{185}}{2*2}=\frac{3-\sqrt{185}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{185}}{2*2}=\frac{3+\sqrt{185}}{4}
$