q2+25q+24=0

Solution for q2+25q+24=0 equation:

q2+25q+24=0

We add all the numbers together, and all the variables

q^2+25q+24=0

a = 1; b = 25; c = +24;
Δ = b2-4ac
Δ = 252-4·1·24
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*1}=\frac{-48}{2}
=-24
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*1}=\frac{-2}{2}
=-1
$