t^2-6t-35=0

Solution for t^2-6t-35=0 equation:

t^2-6t-35=0

a = 1; b = -6; c = -35;
Δ = b2-4ac
Δ = -62-4·1·(-35)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{11}}{2*1}=\frac{6-4\sqrt{11}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{11}}{2*1}=\frac{6+4\sqrt{11}}{2}
$