1v^2-14v+37=0

Solution for 1v^2-14v+37=0 equation:

1v^2-14v+37=0

We add all the numbers together, and all the variables

v^2-14v+37=0

a = 1; b = -14; c = +37;
Δ = b2-4ac
Δ = -142-4·1·37
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{3}}{2*1}=\frac{14-4\sqrt{3}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{3}}{2*1}=\frac{14+4\sqrt{3}}{2}
$