t^2-18t-48=0

Solution for t^2-18t-48=0 equation:

t^2-18t-48=0

a = 1; b = -18; c = -48;
Δ = b2-4ac
Δ = -182-4·1·(-48)
Δ = 516
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{129}}{2*1}=\frac{18-2\sqrt{129}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{129}}{2*1}=\frac{18+2\sqrt{129}}{2}
$