q^2+49q+144=0

Solution for q^2+49q+144=0 equation:

q^2+49q+144=0

a = 1; b = 49; c = +144;
Δ = b2-4ac
Δ = 492-4·1·144
Δ = 1825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-5\sqrt{73}}{2*1}=\frac{-49-5\sqrt{73}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+5\sqrt{73}}{2*1}=\frac{-49+5\sqrt{73}}{2}
$