2v^2-22v+56=0

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Solution for 2v^2-22v+56=0 equation: 



2v^2-22v+56=0

a = 2; b = -22; c = +56;
Δ = b2-4ac
Δ = -222-4·2·56
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-6}{2*2}=\frac{16}{4}
=4
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+6}{2*2}=\frac{28}{4}
=7
$

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