2k^2+25k+12=0

Solution for 2k^2+25k+12=0 equation:

2k^2+25k+12=0

a = 2; b = 25; c = +12;
Δ = b2-4ac
Δ = 252-4·2·12
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*2}=\frac{-48}{4}
=-12
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*2}=\frac{-2}{4}
=-1/2
$