(4x-7)+(2x+3)-8x(x-4)=0

Solution for (4x-7)+(2x+3)-8x(x-4)=0 equation:

(4x-7)+(2x+3)-8x(x-4)=0

We multiply parentheses

-8x^2+(4x-7)+(2x+3)+32x=0

We get rid of parentheses

-8x^2+4x+2x+32x-7+3=0

We add all the numbers together, and all the variables

-8x^2+38x-4=0

a = -8; b = 38; c = -4;
Δ = b2-4ac
Δ = 382-4·(-8)·(-4)
Δ = 1316
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:

$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{329}}{2*-8}=\frac{-38-2\sqrt{329}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{329}}{2*-8}=\frac{-38+2\sqrt{329}}{-16}
$