4k2+19k=150

Solution for 4k2+19k=150 equation:

4k^2+19k=150

We move all terms to the left:

4k^2+19k-(150)=0

a = 4; b = 19; c = -150;
Δ = b2-4ac
Δ = 192-4·4·(-150)
Δ = 2761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{2761}}{2*4}=\frac{-19-\sqrt{2761}}{8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{2761}}{2*4}=\frac{-19+\sqrt{2761}}{8}
$