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

Solution for (2k^2+k)-3=0 equation:

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

We get rid of parentheses

2k^2+k-3=0

a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4}
=-1+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*2}=\frac{4}{4}
=1
$