6k^2-29k+28=-7

Solution for 6k^2-29k+28=-7 equation:

6k^2-29k+28=-7

We move all terms to the left:

6k^2-29k+28-(-7)=0

We add all the numbers together, and all the variables

6k^2-29k+35=0

a = 6; b = -29; c = +35;
Δ = b2-4ac
Δ = -292-4·6·35
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-1}{2*6}=\frac{28}{12}
=2+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+1}{2*6}=\frac{30}{12}
=2+1/2
$