0=2(x)^2-16(x)+29

Solution for 0=2(x)^2-16(x)+29 equation:

0=2(x)^2-16(x)+29

We move all terms to the left:

0-(2(x)^2-16(x)+29)=0

We add all the numbers together, and all the variables

-(2x^2-16x+29)=0

We get rid of parentheses

-2x^2+16x-29=0

a = -2; b = 16; c = -29;
Δ = b2-4ac
Δ = 162-4·(-2)·(-29)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{6}}{2*-2}=\frac{-16-2\sqrt{6}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{6}}{2*-2}=\frac{-16+2\sqrt{6}}{-4}
$