t2+14t+48=0

Solution for t2+14t+48=0 equation:

t2+14t+48=0

We add all the numbers together, and all the variables

t^2+14t+48=0

a = 1; b = 14; c = +48;
Δ = b2-4ac
Δ = 142-4·1·48
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*1}=\frac{-16}{2}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*1}=\frac{-12}{2}
=-6
$