(k+1)(2k-5)=0

Solution for (k+1)(2k-5)=0 equation:

(k+1)(2k-5)=0

We multiply parentheses ..

(+2k^2-5k+2k-5)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

2k^2-3k-5=0

a = 2; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*2}=\frac{-4}{4}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*2}=\frac{10}{4}
=2+1/2
$