2x^2-8x+21=16

Solution for 2x^2-8x+21=16 equation:

2x^2-8x+21=16

We move all terms to the left:

2x^2-8x+21-(16)=0

We add all the numbers together, and all the variables

2x^2-8x+5=0

a = 2; b = -8; c = +5;
Δ = b2-4ac
Δ = -82-4·2·5
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{-(-8)-2\sqrt{6}}{2*2}=\frac{8-2\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{6}}{2*2}=\frac{8+2\sqrt{6}}{4}
$