2k^2+21k+49=0

Solution for 2k^2+21k+49=0 equation:

2k^2+21k+49=0

a = 2; b = 21; c = +49;
Δ = b2-4ac
Δ = 212-4·2·49
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{-(21)-7}{2*2}=\frac{-28}{4}
=-7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+7}{2*2}=\frac{-14}{4}
=-3+1/2
$