(2k-8k2+7)+(5k2+7k)+(3+4k)=0

Solution for (2k-8k2+7)+(5k2+7k)+(3+4k)=0 equation:

(2k-8k^2+7)+(5k^2+7k)+(3+4k)=0

We add all the numbers together, and all the variables

(2k-8k^2+7)+(5k^2+7k)+(4k+3)=0

We get rid of parentheses

-8k^2+5k^2+2k+7k+4k+7+3=0

We add all the numbers together, and all the variables

-3k^2+13k+10=0

a = -3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·(-3)·10
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-3}=\frac{-30}{-6}
=+5
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-3}=\frac{4}{-6}
=-2/3
$