x2+19x+48=0

Solution for x2+19x+48=0 equation:

x2+19x+48=0

We add all the numbers together, and all the variables

x^2+19x+48=0

a = 1; b = 19; c = +48;
Δ = b2-4ac
Δ = 192-4·1·48
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*1}=\frac{-32}{2}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*1}=\frac{-6}{2}
=-3
$