(2(k)^2)-7(k)-1=3

Solution for (2(k)^2)-7(k)-1=3 equation:

(2(k)^2)-7(k)-1=3

We move all terms to the left:

(2(k)^2)-7(k)-1-(3)=0

determiningTheFunctionDomain
2k^2-7k-1-3=0

We add all the numbers together, and all the variables

2k^2-7k-4=0

a = 2; b = -7; c = -4;
Δ = b2-4ac
Δ = -72-4·2·(-4)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*2}=\frac{-2}{4}
=-1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*2}=\frac{16}{4}
=4
$