3k+2k^2-7=-7k+20

Solution for 3k+2k^2-7=-7k+20 equation:

3k+2k^2-7=-7k+20

We move all terms to the left:

3k+2k^2-7-(-7k+20)=0

We get rid of parentheses

2k^2+3k+7k-20-7=0

We add all the numbers together, and all the variables

2k^2+10k-27=0

a = 2; b = 10; c = -27;
Δ = b2-4ac
Δ = 102-4·2·(-27)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{79}}{2*2}=\frac{-10-2\sqrt{79}}{4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{79}}{2*2}=\frac{-10+2\sqrt{79}}{4}
$