1t^2-38t-38=0

Solution for 1t^2-38t-38=0 equation:

1t^2-38t-38=0

We add all the numbers together, and all the variables

t^2-38t-38=0

a = 1; b = -38; c = -38;
Δ = b2-4ac
Δ = -382-4·1·(-38)
Δ = 1596
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:

$\sqrt{\Delta}=\sqrt{1596}=\sqrt{4*399}=\sqrt{4}*\sqrt{399}=2\sqrt{399}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{399}}{2*1}=\frac{38-2\sqrt{399}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{399}}{2*1}=\frac{38+2\sqrt{399}}{2}
$