8x(x-5)+(3x+2)=x-14

Solution for 8x(x-5)+(3x+2)=x-14 equation:

8x(x-5)+(3x+2)=x-14

We move all terms to the left:

8x(x-5)+(3x+2)-(x-14)=0

We multiply parentheses

8x^2-40x+(3x+2)-(x-14)=0

We get rid of parentheses

8x^2-40x+3x-x+2+14=0

We add all the numbers together, and all the variables

8x^2-38x+16=0

a = 8; b = -38; c = +16;
Δ = b2-4ac
Δ = -382-4·8·16
Δ = 932
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:

$\sqrt{\Delta}=\sqrt{932}=\sqrt{4*233}=\sqrt{4}*\sqrt{233}=2\sqrt{233}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{233}}{2*8}=\frac{38-2\sqrt{233}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{233}}{2*8}=\frac{38+2\sqrt{233}}{16}
$