7k^2=6k+3=3

Solution for 7k^2=6k+3=3 equation:

7k^2=6k+3=3

We move all terms to the left:

7k^2-(6k+3)=0

We get rid of parentheses

7k^2-6k-3=0

a = 7; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·7·(-3)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{30}}{2*7}=\frac{6-2\sqrt{30}}{14}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{30}}{2*7}=\frac{6+2\sqrt{30}}{14}
$