4q^2+49q+12=0

Solution for 4q^2+49q+12=0 equation:

4q^2+49q+12=0

a = 4; b = 49; c = +12;
Δ = b2-4ac
Δ = 492-4·4·12
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-47}{2*4}=\frac{-96}{8}
=-12
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+47}{2*4}=\frac{-2}{8}
=-1/4
$