3k+15=2k^2-12k+20

Solution for 3k+15=2k^2-12k+20 equation:

3k+15=2k^2-12k+20

We move all terms to the left:

3k+15-(2k^2-12k+20)=0

We get rid of parentheses

-2k^2+3k+12k-20+15=0

We add all the numbers together, and all the variables

-2k^2+15k-5=0

a = -2; b = 15; c = -5;
Δ = b2-4ac
Δ = 152-4·(-2)·(-5)
Δ = 185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{185}}{2*-2}=\frac{-15-\sqrt{185}}{-4}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{185}}{2*-2}=\frac{-15+\sqrt{185}}{-4}
$