v^2-12v+21=4

Solution for v^2-12v+21=4 equation:

v^2-12v+21=4

We move all terms to the left:

v^2-12v+21-(4)=0

We add all the numbers together, and all the variables

v^2-12v+17=0

a = 1; b = -12; c = +17;
Δ = b2-4ac
Δ = -122-4·1·17
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{19}}{2*1}=\frac{12-2\sqrt{19}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{19}}{2*1}=\frac{12+2\sqrt{19}}{2}
$