3c^2-22c+7=0

Solution for 3c^2-22c+7=0 equation:

3c^2-22c+7=0

a = 3; b = -22; c = +7;
Δ = b2-4ac
Δ = -222-4·3·7
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-20}{2*3}=\frac{2}{6}
=1/3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+20}{2*3}=\frac{42}{6}
=7
$