35k2-22k+7=4

Solution for 35k2-22k+7=4 equation:

35k^2-22k+7=4

We move all terms to the left:

35k^2-22k+7-(4)=0

We add all the numbers together, and all the variables

35k^2-22k+3=0

a = 35; b = -22; c = +3;
Δ = b2-4ac
Δ = -222-4·35·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-8}{2*35}=\frac{14}{70}
=1/5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+8}{2*35}=\frac{30}{70}
=3/7
$